A nucleus disintegrates into two nuclear parts, in such a way that ratio of their nuclear sizes is $1: 2^{1 / 3}$. Their respective speed have a ratio of $n: 1$. The value of $n$ is __________.
Explanation:
$ \frac{m_1}{m_2} = \left(\frac{1}{2^{1/3}}\right)^3 = \frac{1}{2} $
Now, according to the conservation of linear momentum, the momentum before disintegration is equal to the momentum after disintegration:
$ m_1 v_1 = m_2 v_2 $
From the problem statement, the ratio of their respective speeds is $n : 1$, so we can write:
$ v_1 = n \cdot v_2 $
Substitute the expression for $v_1$ into the momentum conservation equation:
$ m_1 (n \cdot v_2) = m_2 v_2 $
We know the mass ratio, so substitute that into the equation:
$ \frac{1}{2} m_2 (n \cdot v_2) = m_2 v_2 $
Divide both sides by $m_2 v_2$:
$ \frac{1}{2} n = 1 $
Now, solve for $n$:
$ n = 2 $
Thus, the value of $n$ is 2.
If 917 $\mathop A\limits^o $ be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be ___________ $\mathop A\limits^o $.
Explanation:
The energy difference formula for transitions between energy levels in a hydrogen atom, which is given by
$ \Delta E = -13.6 \, \text{eV} \times \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) $
where ($n_1$) and ($n_2$) are the initial and final energy levels, respectively. For the Lyman series, the electron transitions to the ground state (($n_1$ = 1)), and for the Balmer series, the electron transitions to the first excited state (($n_1$ = 2)).
From the given information, we have the lowest wavelength in the Lyman series, ($\lambda_1 = 917 \, \text{Å}$). Therefore, the energy difference ($\Delta E$) for the Lyman series is
$ \Delta E = \frac{hc}{\lambda_1} $
where (h) is Planck's constant and (c) is the speed of light.
Similarly, for the Balmer series, the energy difference ($\Delta E$) is
$ \Delta E = -13.6 \, \text{eV} \times \left(\frac{1}{2^2} - \frac{1}{\infty^2}\right) = -13.6 \, \text{eV} \times \frac{1}{4} $
The corresponding wavelength ($\lambda_2$) is
$ \lambda_2 = \frac{hc}{\Delta E} $
By comparing ($\Delta E$) for the Lyman and Balmer series, you found that
$ \frac{\lambda_1}{\lambda_2} = \frac{\Delta E_2}{\Delta E_1} = \frac{1}{4} $
Therefore, the lowest wavelength in the Balmer series is
$ \lambda_2 = 4 \lambda_1 = 4 \times 917 \, \text{Å} = 3668 \, \text{Å} $
The decay constant for a radioactive nuclide is 1.5 $\times$ 10$^{-5}$ s$^{-1}$. Atomic weight of the substance is 60 g mole$^{-1}$, ($N_A=6\times10^{23}$). The activity of 1.0 $\mu$g of the substance is ___________ $\times$ 10$^{10}$ Bq.
Explanation:
The activity of a radioactive substance is defined as the rate of decay or disintegration of the substance. It is given by the following formula:
$A = \lambda N$
where $A$ is the activity, $\lambda$ is the decay constant, and $N$ is the number of radioactive atoms present.
We can use this formula to find the activity of 1.0 $\mu$g (or $10^{-6}$ g) of the substance. First, we need to find the number of radioactive atoms present in 1.0 $\mu$g of the substance. We can use the following formula to do this:
$N = \frac{m}{M} N_A$
where $m$ is the mass of the substance, $M$ is its molar mass, and $N_A$ is Avogadro's number.
Substituting the given values, we get:
$N = \frac{1.0 \times 10^{-6} \, \text{g}}{60 \, \text{g/mol}} \times 6 \times 10^{23} \approx 10^{16} \text{ atoms}$
Now, we can use the formula for activity:
$A = \lambda N = (1.5 \times 10^{-5} \, \text{s}^{-1}) (10^{16}) = 1.5 \times 10^{11} \, \text{Bq}$
Therefore, the activity of 1.0 $\mu$g of the substance is 15 $\times$ 10$^{10}$ Bq.
The ratio of wavelength of spectral lines $\mathrm{H}_{\alpha}$ and $\mathrm{H}_{\beta}$ in the Balmer series is $\frac{x}{20}$. The value of $x$ is _________.
Explanation:
The Balmer series corresponds to electronic transitions in a hydrogen atom that terminate in the second (n=2) energy level. The spectral lines in the Balmer series are often labeled according to a Greek letter scheme, with H$_\alpha$ corresponding to the n=3 to n=2 transition, H$_\beta$ corresponding to the n=4 to n=2 transition, and so on.
The wavelength of a spectral line in the Balmer series can be calculated using the Rydberg formula:
$ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right) $
where $R_H$ is the Rydberg constant for hydrogen, $n$ is the principal quantum number corresponding to the initial energy level, and $\lambda$ is the wavelength of the spectral line.
Using this formula, the wavelength of the H$_\alpha$ line is:
$ \frac{1}{\lambda_{\alpha}} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) $
And the wavelength of the H$_\beta$ line is:
$ \frac{1}{\lambda_{\beta}} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) $
Therefore, the ratio of the wavelengths of the H$_\alpha$ and H$_\beta$ lines is:
$ \frac{\lambda_{\alpha}}{\lambda_{\beta}} = \frac{\left( \frac{1}{2^2} - \frac{1}{4^2} \right)}{\left( \frac{1}{2^2} - \frac{1}{3^2} \right)} = \frac{\frac{3}{16}}{\frac{5}{36}} = \frac{27}{20} $
So, comparing with the given ratio $\frac{x}{20}$, we find that $x = 27$.
A nucleus with mass number 242 and binding energy per nucleon as $7.6~ \mathrm{MeV}$ breaks into two fragment each with mass number 121. If each fragment nucleus has binding energy per nucleon as $8.1 ~\mathrm{MeV}$, the total gain in binding energy is _________ $\mathrm{MeV}$.
Explanation:
The total binding energy of a nucleus is the binding energy per nucleon multiplied by the number of nucleons (protons and neutrons), which is the mass number.
The initial total binding energy of the nucleus is $242 \times 7.6 \, \text{MeV}$.
After the break, each fragment has a total binding energy of $121 \times 8.1 \, \text{MeV}$.
Since there are two such fragments, the final total binding energy is $2 \times 121 \times 8.1 \, \text{MeV}$.
The gain in binding energy is the final total binding energy minus the initial total binding energy. Therefore, the gain in binding energy is:
$2 \times 121 \times 8.1 \, \text{MeV} - 242 \times 7.6 \, \text{MeV} = 1960.2 \, \text{MeV} - 1839.2 \, \text{MeV} = 121 \, \text{MeV}$.
Therefore, the total gain in binding energy is $121 \, \text{MeV}$.
Experimentally it is found that $12.8 ~\mathrm{eV}$ energy is required to separate a hydrogen atom into a proton and an electron. So the orbital radius of the electron in a hydrogen atom is $\frac{9}{x} \times 10^{-10} \mathrm{~m}$. The value of the $x$ is __________.
$\left(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}, \frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right.$ and electronic charge $\left.=1.6 \times 10^{-19} \mathrm{C}\right)$
Explanation:
The binding energy of an electron in a hydrogen atom is given by the formula:
$ E = \frac{k e^2}{2 r} $
where:
- $E$ is the energy of the electron,
- $k$ is Coulomb's constant ($9 \times 10^9 \, \text{Nm}^2/\text{C}^2$),
- $e$ is the charge of the electron ($1.6 \times 10^{-19} \, \text{C}$), and
- $r$ is the radius of the orbit.
In this scenario, the energy $E$ required to separate a hydrogen atom into a proton and an electron is given as $12.8 \, \text{eV}$, which needs to be converted into joules using the conversion factor $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$. So,
$ 12.8 \, \text{eV} = 12.8 \times 1.6 \times 10^{-19} \, \text{J} $
We can then substitute the given values into the energy equation and solve for $r$:
$ 12.8 \times 1.6 \times 10^{-19} \, \text{J} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{2r} $
Solving for $r$, we get:
$ r = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{2 \times 12.8 \times 1.6 \times 10^{-19}} $
This simplifies to:
$ r = \frac{9 \times 10^{-10}}{16} $
Comparing this with the given form of the radius, which is $\frac{9}{x} \times 10^{-10}$, we find that the value of $x$ is 16.
The radius of fifth orbit of the $\mathrm{Li}^{++}$ is __________ $\times 10^{-12} \mathrm{~m}$.
Take: radius of hydrogen atom $ = 0.51\,\mathop A\limits^o $
Explanation:
The formula to calculate the radius of an orbit for a hydrogen-like atom/ion is:
$ r_n = r_0 \frac{n^2}{Z} $
where:
- $r_n$ is the radius of the nth orbit,
- $n$ is the principal quantum number (the orbit number),
- $r_0$ is the Bohr radius (radius of the first Bohr orbit in the hydrogen atom), and
- $Z$ is the atomic number (the number of protons in the nucleus).
We're dealing with a Li²⁺ ion and we're interested in the fifth orbit ($n = 5$), and given that $r_0$ is 0.51 Å and $Z$ for Li is 3, we can substitute these values into the formula:
$ r_5 = 0.51 \times \frac{25}{3} \text{ Å} = 4.25 \text{ Å} $
which is $4.25 \times 10^{-10}$ m, or equivalently $425 \times 10^{-12}$ m when converted to meters.
Nucleus A having $Z=17$ and equal number of protons and neutrons has $1.2 ~\mathrm{MeV}$ binding energy per nucleon.
Another nucleus $\mathrm{B}$ of $Z=12$ has total 26 nucleons and $1.8 ~\mathrm{MeV}$ binding energy per nucleons.
The difference of binding energy of $\mathrm{B}$ and $\mathrm{A}$ will be _____________ $\mathrm{MeV}$.
Explanation:
$ \begin{aligned} & \mathrm{Z}=17=\text { Number of protons } \\\\ & Given, Z = N \\\\ & \therefore N = 17 \\\\ & A=34=Z+N \\\\ & E_{b n}=1.2 \mathrm{MeV} \\\\ & \frac{\left(E_B\right)_1}{A}=1.2 \mathrm{MeV} \\\\ & \left(\mathrm{E}_{\mathrm{B}}\right)_1=(1.2 \mathrm{MeV}) \times \mathrm{A} \\\\ & \left(\mathrm{E}_{\mathrm{B}}\right)_1=(1.2 \mathrm{MeV}) \times 34 \\\\ & \left(\mathrm{E}_{\mathrm{B}}\right)_1=40.8 \mathrm{MeV} \\\\ & \end{aligned} $
For Nucleus B :
$ \begin{aligned} & \mathrm{Z}=12, \mathrm{~A}=26 \\\\ & \mathrm{E}_{\mathrm{bn}}=1.8 \mathrm{MeV} \\\\ & \frac{\left(\mathrm{E}_{\mathrm{b}}\right)_2}{\mathrm{~A}}=1.8 \mathrm{MeV} \\\\ & \left(\mathrm{E}_{\mathrm{b}}\right)_2=(1.8 \mathrm{MeV}) \times \mathrm{A} \\\\ & \left(\mathrm{E}_{\mathrm{b}}\right)_2=(1.8 \mathrm{MeV}) \times 26 \\\\ & \left(\mathrm{E}_{\mathrm{b}}\right)_2=46.8 \mathrm{MeV} \end{aligned} $
Therefore, difference in binding energy of $\mathrm{B}$ and $\mathrm{A}$ is
$ \begin{aligned} \Delta \mathrm{E}_{\mathrm{b}} & =\left(\mathrm{E}_{\mathrm{b}}\right)_2-\left(\mathrm{E}_{\mathrm{b}}\right)_2 \\\\ & =46.8 \mathrm{MeV}-40.8 \mathrm{MeV}=6 \mathrm{MeV} \end{aligned} $
A light of energy $12.75 ~\mathrm{eV}$ is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is $\frac{x}{\pi} \times 10^{-17} ~\mathrm{eVs}$. The value of $x$ is ___________ (use $h=4.14 \times 10^{-15} ~\mathrm{eVs}, c=3 \times 10^{8} \mathrm{~ms}^{-1}$ ).
Explanation:
$ \begin{aligned} & 12.75=13.6\left[\frac{1}{1^{2}}-\frac{1}{n^{2}}\right] \\\\ & \Rightarrow n=4 \\\\ & \text { So, Angular momentum } L=\frac{n h}{2 \pi}=\frac{2 h}{\pi} \end{aligned} $
$ \begin{aligned} \text { Angular momentum }=\frac{2}{\pi} & \times 4.14 \times 10^{-15} \\\\ & =\frac{828 \times 10^{-17}}{\pi} \mathrm{eVs} \end{aligned} $
Explanation:
$ \begin{aligned} & \mathrm{E}_{\mathrm{Li}^{2+}}=13.6 \frac{Z^{2}}{n^{2}}=13.6 \times \frac{9}{9}=13.6 \mathrm{eV} \\\\ & =136 \times 10^{-1} \mathrm{eV} \end{aligned} $
For hydrogen atom, $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of $\lambda_{1}$ and $\lambda_{2}$ is $\frac{x}{32}$. The value of $x$ is __________.
Explanation:
$ \begin{aligned} & \frac{1}{\lambda_{1}}=\mathrm{RZ}^{2}\left[\frac{1}{1^{2}}-\frac{1}{3^{2}}\right]=\frac{8}{9} \mathrm{RZ}^{2} ........(1)\\\\ & \frac{1}{\lambda_{2}}=\mathrm{RZ}^{2}\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]=\frac{3}{4} \mathrm{RZ}^{2} ........(2) \end{aligned} $
$ \begin{gathered} 1 / 2 \Rightarrow \frac{\lambda_{2}}{\lambda_{1}}=\frac{8}{9} \times \frac{4}{3}=\frac{32}{27} \\\\ \frac{\lambda_{1}}{\lambda_{2}}=\frac{27}{32} \end{gathered} $
A radioactive nucleus decays by two different process. The half life of the first process is 5 minutes and that of the second process is $30 \mathrm{~s}$. The effective half-life of the nucleus is calculated to be $\frac{\alpha}{11} \mathrm{~s}$. The value of $\alpha$ is __________.
Explanation:
$ \Rightarrow {\lambda _{eff}} = {\lambda _1} + {\lambda _2}$
$ \Rightarrow {{\ln 2} \over {{t_{1/2}}}} = {{\ln 2} \over {{{({t_{1/2}})}_1}}} + {{\ln 2} \over {{{({t_{1/2}})}_2}}}$
$ \Rightarrow {t_{1/2}} = {{{{({t_{1/2}})}_1} \times {{({t_{1/2}})}_2}} \over {{{({t_{1/2}})}_1} + {{({t_{1/2}})}_2}}} = {{300 \times 30} \over {300 + 30}}s = {{300} \over {11}}s$
$ \Rightarrow \alpha = 300$
A radioactive element $_{92}^{242}$X emits two $\alpha$-particles, one electron and two positrons. The product nucleus is represented by $_{\mathrm{P}}^{234}$Y. The value of P is __________.
Explanation:
So, $P=87$
A nucleus disintegrates into two smaller parts, which have their velocities in the ratio 3 : 2. The ratio of their nuclear sizes will be ${\left( {{x \over 3}} \right)^{{1 \over 3}}}$. The value of '$x$' is :-
Explanation:
Since, Nuclear mass density is constant
$ \begin{aligned} & \frac{\mathrm{m}_1}{\frac{4}{3} \pi \mathrm{r}_1^3}=\frac{\mathrm{m}_2}{\frac{4}{3} \pi \mathrm{r}_2^3} \\\\ & \left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^3=\frac{\mathrm{m}_1}{\mathrm{~m}_2} \\\\ & \frac{\mathrm{r}_1}{\mathrm{r}_2}=\left(\frac{2}{3}\right)^{\frac{1}{3}} \\\\ & \text { So, } \mathrm{x}=2 \end{aligned} $
The wavelength of the radiation emitted is $\lambda_0$ when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will $\frac{20}{x}\lambda_0$. The value of $x$ is _____________.
Explanation:
$\frac{1}{\lambda_{0}}=R\left(\frac{1}{4}-\frac{1}{9}\right)=\left(\frac{5 R}{36}\right)$
Third excited state $ \to $ second orbit , $n=4$ to $n=2$
$\frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{16}\right)=\left(\frac{3}{16} R\right)$
Taking ratio of (1) and (2)
$\frac{\lambda}{\lambda_{0}}=\frac{5}{36} \times \frac{16}{3}=\left(\frac{20}{27}\right)$
$\lambda=\frac{20}{27} \lambda_{0}$
$x=27$
The energy released per fission of nucleus of $^{240}$X is 200 MeV. The energy released if all the atoms in 120g of pure $^{240}$X undergo fission is ____________ $\times$ 10$^{25}$ MeV.
(Given $\mathrm{N_A=6\times10^{23}}$)
Explanation:
Number of atom of $X=\frac{1}{2} \times N_{A}=3 \times 10^{23}$ atom
Energy released $=3 \times 10^{23} \times 200 ~ \mathrm{MeV}$
$ =6 \times 10^{25} ~\mathrm{MeV} $
Assume that protons and neutrons have equal masses. Mass of a nucleon is $1.6\times10^{-27}$ kg and radius of nucleus is $1.5\times10^{-15}~\mathrm{A^{1/3}}$ m. The approximate ratio of the nuclear density and water density is $n\times10^{13}$. The value of $n$ is __________.
Explanation:
$ \text { Volume }=\frac{4 \pi}{3} r^{3} $
Mass of nucleus $=\left(1.6 \times 10^{-27}\right) \mathrm{A} \mathrm{kg}$
$ \text { Density of nucleus }=\frac{1.6 \times 10^{-27} \times A}{\frac{4}{3} \times \pi \times\left(1.5 \times 10^{-15} A^{\frac{1}{3}}\right)^{3}} $
$ \begin{aligned} & =\frac{1.6 \times 3 \times 8 \times 10^{18}}{4 \pi \times 27} \\\\ & =\frac{32}{9 \pi} \times 10^{17} \end{aligned} $
Density of water $=1000 \mathrm{~kg} / \mathrm{m}^{3}$
$\frac{\text { Density of nucleus }}{\text { Density of water }}=\frac{\frac{32}{9 \pi} \times 10^{17}}{1000}$
$=\frac{320}{9 \pi} \times 10^{13}$
$=11.32 \times 10^{13}$
value of $n=11$
Two radioactive materials A and B have decay constants $25 \lambda$ and $16 \lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of B to that of A will be "e" after a time $\frac{1}{a \lambda}$. The value of a is _________.
Explanation:
$N_{B}=N_{0} e^{-16 \lambda t}$
$\frac{N_{B}}{N_{A}}=e=e^{9 \lambda t}$
$t=\frac{1}{9 \lambda}$
A freshly prepared radioactive source of half life 2 hours 30 minutes emits radiation which is 64 times the permissible safe level. The minimum time, after which it would be possible to work safely with source, will be _________ hours.
Explanation:
${T_{1/2}} = 150$ minutes
${A_0} = 64x$, where x is safe limit
$x = 64x \times {2^{ - {n \over {{T_{1/2}}}}}}$
$ \Rightarrow {1 \over {64}} = {2^{ - {n \over {{T_{1/2}}}}}}$
or ${n \over {{T_{1/2}}}} = 6$
$ \Rightarrow n = 6 \times 150$ minutes
= 15 hours
Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below :
${ }_{1}^{2} X+{ }_{1}^{2} X={ }_{2}^{4} Y$
The binding energies per nucleon for $\frac{2}{1} X$ and ${ }_{2}^{4} Y$ are $1.1 \,\mathrm{MeV}$ and $7.6 \,\mathrm{MeV}$ respectively. The energy released in this process is _______________ $\mathrm{MeV}$.
Explanation:
Energy released = Change in B.E.
(7.6 $\times$ 4) $-$ [4 $\times$ 1.1] = 26 MeV
In the hydrogen spectrum, $\lambda$ be the wavelength of first transition line of Lyman series. The wavelength difference will be "a$\lambda$'' between the wavelength of $3^{\text {rd }}$ transition line of Paschen series and that of $2^{\text {nd }}$ transition line of Balmer series where $\mathrm{a}=$ ___________.
Explanation:
${1 \over \lambda } = {R_H}\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$
${1 \over {{\lambda _3}}} = {R_H}\left( {{1 \over {{3^2}}} - {1 \over {{6^2}}}} \right)$
${1 \over {{\lambda _2}}} = {R_H}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right)$
$\therefore$ ${\lambda _3} - {\lambda _2} = a\lambda $
$a = 5$
${x \over {x + 4}}$ is the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its
(i) third permitted energy level to the second level and
(ii) the highest permitted energy level to the second permitted level.
The value of x will be ____________.
Explanation:
${E_n} = - {{13.6} \over {{n^2}}}\,eV$
${{{1 \over {{2^2}}} - {1 \over {{3^2}}}} \over {{1 \over {{2^2}}}}} = {x \over {x + 4}}$
$ \Rightarrow {{9 - 4} \over {9 \times 4 \times {1 \over 4}}} = {x \over {x + 4}} = {5 \over 9}$
$x = 5$
A hydrogen atom in its first excited state absorbs a photon of energy x $\times$ 10$-$2 eV and excited to a higher energy state where the potential energy of electron is $-$1.08 eV. The value of x is ______________.
Explanation:
$ \begin{aligned} & \text { So, } \Delta \mathrm{E}, \mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{i}}=-0.544-\left(-\frac{13.6}{2^{2}}\right)=3.4-0.544 \\\\ & \approx 2.86 \mathrm{eV}=286 \times 10^{-2} \mathrm{eV} \end{aligned} $
The half life of a radioactive substance is 5 years. After x years a given sample of the radioactive substance gets reduced to 6.25% of its initial value. The value of x is ____________.
Explanation:
$N = {N_0}{e^{ - \lambda t}}$
$ \Rightarrow {{6.25} \over {100}} = {e^{ - \lambda t}}$
$ \Rightarrow {e^{ - \lambda t}} = {1 \over {16}} = {\left( {{1 \over 2}} \right)^4}$
$ \Rightarrow t = 4{t_{1/2}}$
$ \Rightarrow t = 20$ years
$\sqrt {{d_1}} $ and $\sqrt {{d_2}} $ are the impact parameters corresponding to scattering angles 60$^\circ$ and 90$^\circ$ respectively, when an $\alpha$ particle is approaching a gold nucleus. For d1 = x d2, the value of x will be ____________.
Explanation:
Impact parameter $\propto$ $\cot {\theta \over 2}$
$ \Rightarrow \sqrt {{{{d_1}} \over {{d_2}}}} = {{\sqrt 3 } \over 1}$
$ \Rightarrow {d_1} = 3{d_2}$
$ \Rightarrow x = 3$
A beam of monochromatic light is used to excite the electron in Li+ + from the first orbit to the third orbit. The wavelength of monochromatic light is found to be x $\times$ 10$-$10 m. The value of x is ___________.
[Given hc = 1242 eV nm]
Explanation:
E(in eV) = 13.6 $\times$ 9$\left( {1 - {1 \over 9}} \right)$
= 13.6 $\times$ 8 eV
$\Rightarrow$ $\lambda = {{12420} \over {13.6 \times 8}}\mathop A\limits^o $
= 114.15 $\mathop A\limits^o $
A sample contains 10$-$2 kg each of two substances A and B with half lives 4 s and 8 s respectively. The ratio of their atomic weights is 1 : 2. The ratio of the amounts of A and B after 16 s is ${x \over {100}}$. The value of x is ___________.
Explanation:
${N_1} = {{\left( {{{{{10}^{ - 2}}} \over 1}} \right)} \over {{2^4}}}$
${N_2} = {{\left( {{{{{10}^{ - 2}}} \over 2}} \right)} \over {{2^2}}}$
$ \Rightarrow {{{N_1}} \over {{N_2}}} = {1 \over 2}$
$\therefore$ Mass ratio of A and B,
${{{m_1}} \over {{m_2}}} = {{{N_1}} \over {{N_2}}} \times \left( {{{{M_1}} \over {{M_2}}}} \right)$
$ = {1 \over 2} \times \left( {{1 \over 2}} \right)$
$ = {1 \over 4}$
$ = {{25} \over {100}}$
$\therefore$ $x = 25$
Explanation:
$ = {{6 \times (6 - 1)} \over 2} = {{6 \times 5} \over 2} = 15$
[h = 4.14 $\times$ 10$-$15 eVs, c = 3 $\times$ 108 ms$-$1]
Explanation:
${{hc} \over {{\lambda _{{k_\alpha }}}}} = {E_k} - {E_L}$
${E_L} = {E_k} - {{hc} \over {{\lambda _{{k_\alpha }}}}}$
= 27.5 KeV $ - {{12.42 \times {{10}^{ - 7}}eVm} \over {0.071 \times {{10}^{ - 9}}m}}$
EL = (27.5 $-$ 17.5) keV
= 10 keV
Explanation:
$I = {{1.6 \times {{10}^{ - 19}} \times 2.2 \times {{10}^6} \times 7} \over {2 \times 22 \times 0.5 \times {{10}^{ - 10}}}}$
= 1.12 mA
= 112 $\times$ 10$-$2 mA
Explanation:
C = 200 $\mu$F
${q \over N} = {{{Q_0}{e^{ - t/RC}}} \over {{N_0}{e^{ - \lambda t}}}} = {{{Q_0}} \over {{N_0}}}{e^{t\left( {\lambda - {1 \over {RC}}} \right)}}$
Since q/N is constant hence
$\lambda ={1 \over {RC}}$
$R = {1 \over {\lambda C}} = {{{T_m}} \over C} = {{30 \times {{10}^{ - 3}}} \over {200 \times {{10}^{ - 6}}}} = 150\Omega $
Mass of neutron = 1.00866 u
Mass of proton = 1.00726 u
Mass of Aluminium nucleus = 27.18846 u
(Assume 1 u corresponds to x J of energy)
(Round off to the nearest integer)
Explanation:
= (13 $\times$ 1.00726 + 14 $\times$ 1.00866) $-$ 27.18846
= 27.21562 $-$ 27.18846
= 0.02716 u
E = 27.16 x $\times$ 10$-$3 J
Explanation:
For half life
${{{A_0}} \over 2} = {e^{ - \lambda {t_{1/2}}}}$
$ \Rightarrow $ ${\lambda {t_{1/2}}}$ = ln 2 .....(1)
And when radioactive element becomes ${\left( {{1 \over 8}} \right)^{th}}$ of its initial value in 30 years
${{{A_0}} \over 8} = {A_0}{e^{ - \lambda \times 30}} \Rightarrow \lambda \times 30 = \ln 8$
$ \Rightarrow $ 30$\lambda = 3\ln 2$
$ \Rightarrow $ $\lambda = {{3\ln 2} \over {30}}$ .....(2)
Putting value of $\lambda $ in (1), we get
${{3\ln 2} \over {30}} \times {t_{1/2}}$ = ln 2
$ \Rightarrow $ ${t_{1/2}}$ = 10 years
Explanation:
$4 \times {t_{1/2}} = 80$
${t_{1/2}} = 20$ days
central potential field U(r) = U0r4. If Bohr's quantization conditions
are applied, radii of possible
orbitals rn vary with ${n^{{1 \over \alpha }}}$, where $\alpha$ is ____________.
Explanation:
$ = - {d \over {dr}}({U_0}{r^4})$
$\overrightarrow F = - 4{U_0}{r^3}$
$ \because $ ${{m{v^2}} \over r} = 4{U_0}{r^3}$
$m{v^2} = 4{U_0}{r^4}$
Then $v \propto {r^2}$
$ \because $ $mvr = {{nh} \over {2\pi }}$
Then ${r^3}\propto\,n$
$r\,\propto \,{(n)^{{1 \over 3}}}$
So the value of $\alpha = 3$
given $\lambda$1, $\lambda$2, $\lambda$3 considering the Bohr atomic model, the wave lengths of first and third spectral lines $\left( \frac{\lambda_{1} }{\lambda_{3} } \right) $ are related by a factor of approximately 'x' $\times$ 10$-$1.
The value of x, to the nearest integer, is _________.
Explanation:
${1 \over {{\lambda _1}}} = R{z^2}\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right)$
${1 \over {{\lambda _1}}} = R{z^2}{5 \over {36}}$ ..... (i)
For 3rd line
${1 \over {{\lambda _3}}} = R{z^2}\left( {{1 \over {{2^2}}} - {1 \over {{5^2}}}} \right)$
${1 \over {{\lambda _3}}} = R{z^2}{{21} \over {100}}$ ...... (ii)
Dividing (ii) by (i),
${{{\lambda _1}} \over {{\lambda _3}}} = {{21} \over {100}} \times {{36} \over 5} = 1.512 = 15.12 \times {10^{ - 1}}$
$x \approx 15$
${N \over 4}$. The value of N is :
(Given the mass of the hydrogen atom to be 1 GeV/c2) ______ .
Explanation:
and mH = 1 GeV/c2 = 1000 MeV/c2 = 5 $ \times $ 200 = 5m
Applying momentum conservation,
pi = pf
$ \Rightarrow $ mv0 + 0 = 0 + 5 mv'
$ \Rightarrow $ v' = ${{{v_0}} \over 5}$
Initial kinetic energy, ki = ${1 \over 2}mv_0^2$
Final kinetic energy, kf = ${1 \over 2}\left( {5m} \right){\left( {{{{v_0}} \over 5}} \right)^2}$
$ \therefore $ Loss in KE
= ${1 \over 2}mv_0^2$ - ${1 \over 2}\left( {5m} \right){\left( {{{{v_0}} \over 5}} \right)^2}$
= ${4 \over 5}\left( {{1 \over 2}mv_0^2} \right)$ = ${4 \over 5}\left( {{k_i}} \right)$
This lost energy is used by the hydrogen atom to move from ground state to the first excited state. We know the the energy required by the hydrogen atom to move from ground state to first excited state is 10.2 eV.
$ \therefore $ ${4 \over 5}\left( {{k_i}} \right)$ = 10.2
$ \Rightarrow $ ki = ${{5 \times 10.2} \over 4} = {{51} \over 4}$
$ \therefore $ N = 51
Explanation:
${1 \over {{\lambda _2}}} = R{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right)$ = ${{12} \over {64}}$RZ2
$ \therefore $ ${{{\lambda _2}} \over {{\lambda _1}}}$ = ${5 \over {36}} \times {{64} \over {12}}$ = ${{20} \over {27}}$
$ \Rightarrow $ $\lambda $2 = ${{20} \over {27}}{\lambda _1}$
= ${{20} \over {27}}$ $ \times $ 6561 = 4860 $\mathop A\limits^o $ = 486 nm
List-I shows various functional dependencies of energy $(E)$ on the atomic number $(Z)$. Energies associated with certain phenomena are given in List-II.
Choose the option that describes the correct match between the entries in List-I to those in List-II.
| List–I | List–II |
|---|---|
| (P) $E \propto Z^2$ | (1) energy of characteristic x-rays |
| (Q) $E \propto (Z - 1)^2$ | (2) electrostatic part of the nuclear binding energy for stable nuclei with mass numbers in the range 30 to 170 |
| (R) $E \propto Z(Z - 1)$ | (3) energy of continuous x-rays |
| (S) $E$ is practically independent of $Z$ | (4) average nuclear binding energy per nucleon for stable nuclei with mass number in the range 30 to 170 |
| (5) energy of radiation due to electronic transitions from hydrogen-like atoms |
P→4, Q→3, R→1, S→2
P→5, Q→2, R→1, S→4
P→5, Q→1, R→2, S→4
P→3, Q→2, R→1, S→5
| List - I | List - II |
|---|---|
| (P) ${ }_{92}^{238} U \rightarrow{ }_{91}^{234} \mathrm{~Pa}$ | (1) one $\alpha$ particle and one $\beta^{+}$particle |
| (Q) ${ }_{82}^{214} \mathrm{~Pb} \rightarrow{ }_{82}^{210} \mathrm{~Pb}$ | (2) three $\beta^{-}$particles and one $\alpha$ particle |
| (R) ${ }_{81}^{210} \mathrm{Tl} \rightarrow{ }_{82}^{206} \mathrm{~Pb}$ | (3) two $\beta^{-}$particles and one $\alpha$ particle |
| (S) ${ }_{91}^{228} \mathrm{~Pa} \rightarrow{ }_{88}^{224} \mathrm{Ra}$ | (4) one $\alpha$ particle and one $\beta^{-}$particle |
| (5) one $\alpha$ particle and two $\beta^{+}$particles |
[Given : In 10 = 2.3]
The measured masses of the neutron, $_1^1H$, $_7^{15}N$ and $_8^{15}O$ are 1.008665u, 1.007825u, 15.000109u and 15.003065u, respectively. Given that the radii of both the $_7^{15}N$ and $_8^{15}O$ nuclei are same, 1 u = 931.5 MeV/c2 (c is the speed of light) and e2/(4$\pi$${{\varepsilon _0}}$) = 1.44 MeV fm. Assuming that the difference between the binding energies of $_7^{15}N$ and $_8^{15}O$ is purely due to the electrostatic energy, the radius of either of the nuclei is (1 fm = 10$-$15 m)
Match the nuclear processes given in Column I with the appropriate option(s) in Column II:
If $\lambda$Cu is the wavelength of K$\alpha$ X-ray line of copper (atomic number 29) and $\lambda$Mo is the wavelength of the K$\alpha$ X-ray line of molybdenum (atomic number 42), then the ratio $\lambda$Cu/$\lambda$Mo is close to
The correct statement is
The kinetic energy (in keV) of the alpha particle, when the nucleus $_{84}^{210}Po$ at rest undergoes alpha decay, is
Match List I of the nuclear processes with List II containing parent nucleus and one of the end products of each process and then select the correct answer using the codes given below the lists :
| List I | List II | ||
|---|---|---|---|
| P. | Alpha decay | 1. | $_8^{15}O \to _7^{15}N + ...$ |
| Q. | ${\beta ^ + }$ decay | 2. | $_{91}^{238}U \to _{90}^{234}Th + ...$ |
| R. | Fission | 3. | $_{83}^{185}Bi \to _{82}^{184}Pb + ...$ |
| S. | Proton emission | 4. | $_{94}^{239}Pu \to _{57}^{140}La + ...$ |
What is the maximum energy of the anti-neutrino?