Atoms and Nuclei
A free neutron decays into a proton but a free proton does not decay into neutron. This is because
Assertion A: The nuclear density of nuclides ${ }_{5}^{10} \mathrm{~B},{ }_{3}^{6} \mathrm{Li},{ }_{26}^{56} \mathrm{Fe},{ }_{10}^{20} \mathrm{Ne}$ and ${ }_{83}^{209} \mathrm{Bi}$ can be arranged as $\rho_{\mathrm{Bi}}^{\mathrm{N}}>\rho_{\mathrm{Fe}}^{\mathrm{N}}>\rho_{\mathrm{Ne}}^{\mathrm{N}}>\rho_{\mathrm{B}}^{\mathrm{N}}>\rho_{\mathrm{Li}}^{\mathrm{N}}$
Reason R: The radius $R$ of nucleus is related to its mass number $A$ as $R=R_{0} A^{1 / 3}$, where $R_{0}$ is a constant.
In the light of the above statements, choose the correct answer from the options given below
Speed of an electron in Bohr's $7^{\text {th }}$ orbit for Hydrogen atom is $3.6 \times 10^{6} \mathrm{~m} / \mathrm{s}$. The corresponding speed of the electron in $3^{\text {rd }}$ orbit, in $\mathrm{m} / \mathrm{s}$ is :
Substance A has atomic mass number 16 and half life of 1 day. Another substance B has atomic mass number 32 and half life of $\frac{1}{2}$ day. If both A and B simultaneously start undergo radio activity at the same time with initial mass 320 g each, how many total atoms of A and B combined would be left after 2 days.
If a radioactive element having half-life of $30 \mathrm{~min}$ is undergoing beta decay, the fraction of radioactive element remains undecayed after $90 \mathrm{~min}$. will be
The energy levels of an atom is shown in figure.
Which one of these transitions will result in the emission of a photon of wavelength 124.1 nm?
Given (h = 6.62 $\times$ 10$^{-34}$ Js)
The ratio of the density of oxygen nucleus ($_8^{16}O$) and helium nucleus ($_2^{4}\mathrm{He}$) is
A photon is emitted in transition from n = 4 to n = 1 level in hydrogen atom. The corresponding wavelength for this transition is (given, h = 4 $\times$ 10$^{-15}$ eVs) :
Consider the following radioactive decay process
$_{84}^{218}A\buildrel \alpha \over \longrightarrow {A_1}\buildrel {{\beta ^ - }} \over \longrightarrow {A_2}\buildrel \gamma \over \longrightarrow {A_3}\buildrel \alpha \over \longrightarrow {A_4}\buildrel {{\beta ^ + }} \over \longrightarrow {A_5}\buildrel \gamma \over \longrightarrow {A_6}$
The mass number and the atomic number of A$_6$ are given by :
Explanation:
$ \begin{aligned} & \text { As } \frac{1}{\lambda}=\mathrm{RZ}^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\\\ & \frac{1}{\lambda_1}=\mathrm{R}(1)^2\left[\frac{1}{(2)^2}-\frac{1}{(3)^2}\right]=\mathrm{R}\left(\frac{5}{36}\right)~~......(i) \\\\ & \& \frac{1}{\lambda_2}=\mathrm{R}(1)^2\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right]=\mathrm{R}\left(\frac{7}{144}\right) ~~......(ii) \end{aligned} $
(ii) $\div$ (i) gives
$ \begin{aligned} & \frac{\lambda_1}{\lambda_2}=\frac{7 / 144}{5 / 36}=\frac{7}{20}=\frac{7}{4 \times 5} \\\\ & \therefore \mathrm{n}=5 \end{aligned} $
The radius of $2^{\text {nd }}$ orbit of $\mathrm{He}^{+}$ of Bohr's model is $r_{1}$ and that of fourth orbit of $\mathrm{Be}^{3+}$ is represented as $r_{2}$. Now the ratio $\frac{r_{2}}{r_{1}}$ is $x: 1$. The value of $x$ is ___________.
Explanation:
To find the value of $x$, we need to first determine the expressions for the radii of the specified orbits for $\mathrm{He}^{+}$ and $\mathrm{Be}^{3+}$ according to Bohr's model. The radius of an orbit in a hydrogen-like atom (an atom with only one electron) is given by:
$r_n = \frac{n^2 \cdot h^2 \cdot \epsilon_0}{\pi \cdot Z \cdot e^2 \cdot m_e}$Where:
- $r_n$ is the radius of the nth orbit
- $n$ is the principal quantum number (orbit number)
- $h$ is the Planck's constant
- $\epsilon_0$ is the vacuum permittivity
- $Z$ is the atomic number (number of protons in the nucleus)
- $e$ is the elementary charge
- $m_e$ is the mass of the electron
- $\pi$ is the mathematical constant pi
In this problem, we are looking at the 2nd orbit of $\mathrm{He}^{+}$ (which has an atomic number $Z = 2$) and the 4th orbit of $\mathrm{Be}^{3+}$ (which has an atomic number $Z = 4$). Let's calculate the radii for these orbits:
For the 2nd orbit of $\mathrm{He}^{+}$ ($n_1 = 2$ and $Z_1 = 2$):
$r_{1} = \frac{n_1^2 \cdot h^2 \cdot \epsilon_0}{\pi \cdot Z_1 \cdot e^2 \cdot m_e}$For the 4th orbit of $\mathrm{Be}^{3+}$ ($n_2 = 4$ and $Z_2 = 4$):
$r_{2} = \frac{n_2^2 \cdot h^2 \cdot \epsilon_0}{\pi \cdot Z_2 \cdot e^2 \cdot m_e}$We are asked to find the ratio $\frac{r_{2}}{r_{1}}$, which is equal to $x: 1$:
$\frac{r_{2}}{r_{1}} = \frac{\frac{n_2^2 \cdot h^2 \cdot \epsilon_0}{\pi \cdot Z_2 \cdot e^2 \cdot m_e}}{\frac{n_1^2 \cdot h^2 \cdot \epsilon_0}{\pi \cdot Z_1 \cdot e^2 \cdot m_e}}$By simplifying the expression, we get:
$\frac{r_{2}}{r_{1}} = \frac{n_2^2 \cdot Z_1}{n_1^2 \cdot Z_2} = \frac{4^2 \cdot 2}{2^2 \cdot 4}$Now we can calculate the value of $x$:
$x = \frac{r_{2}}{r_{1}} = \frac{16 \cdot 2}{4 \cdot 4} = \frac{32}{16} = 2$Therefore, the value of $x$ in the ratio $\frac{r_{2}}{r_{1}} = x: 1$ is $\boxed{2}$.
A common example of alpha decay is ${ }_{92}^{238} \mathrm{U} \longrightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2} \mathrm{He}^{4}+\mathrm{Q}$
Given :
${ }_{92}^{238} \mathrm{U}=238.05060 ~\mathrm{u}$,
${ }_{90}^{234} \mathrm{Th}=234.04360 ~\mathrm{u}$,
${ }_{2}^{4} \mathrm{He}=4.00260 ~\mathrm{u}$ and
$1 \mathrm{u}=931.5 \frac{\mathrm{MeV}}{c^{2}}$
The energy released $(Q)$ during the alpha decay of ${ }_{92}^{238} \mathrm{U}$ is __________ MeV
Explanation:
Mass difference = Mass of Uranium-238 - (Mass of Thorium-234 + Mass of Helium-4)
Mass difference = $238.05060 \mathrm{u} - (234.04360 \mathrm{u} + 4.00260 \mathrm{u})$
Mass difference = $238.05060 \mathrm{u} - 238.04620 \mathrm{u}$
Mass difference = $0.00440 \mathrm{u}$
Now, we can convert this mass difference to energy using the given conversion factor:
Energy released (Q) = Mass difference × $\frac{931.5 \mathrm{MeV}}{c^2}$
Q = $0.00440 \mathrm{u} × 931.5 \frac{\mathrm{MeV}}{c^2}$
Q ≈ 4.1 MeV
The energy released (Q) during the alpha decay of $^{238}U$ is approximately 4.1 MeV.
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 $n=11$
Read the following statements :
(A) Volume of the nucleus is directly proportional to the mass number.
(B) Volume of the nucleus is independent of mass number.
(C) Density of the nucleus is directly proportional to the mass number.
(D) Density of the nucleus is directly proportional to the cube root of the mass number.
(E) Density of the nucleus is independent of the mass number.
Choose the correct option from the following options.
Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.
A radioactive sample decays $\frac{7}{8}$ times its original quantity in 15 minutes. The half-life of the sample is
The half life period of a radioactive substance is 60 days. The time taken for $\frac{7}{8}$th of its original mass to disintegrate will be :
The activity of a radioactive material is $6.4 \times 10^{-4}$ curie. Its half life is 5 days. The activity will become $5 \times 10^{-6}$ curie after :
What is the half-life period of a radioactive material if its activity drops to $1 / 16^{\text {th }}$ of its initial value in 30 years?
A nucleus of mass $M$ at rest splits into two parts having masses $\frac{M^{\prime}}{3}$ and ${{2M'} \over 3}(M' < M)$. The ratio of de Broglie wavelength of two parts will be :
Mass numbers of two nuclei are in the ratio of $4: 3$. Their nuclear densities will be in the ratio of
The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minute. 10 minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is :
$\left(\right.$Take $\left.\log _{10} 1.88=0.274\right)$
Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength $\lambda$. The value of principal quantum number '$n$' of the excited state will be : ($\mathrm{R}:$ Rydberg constant)
The momentum of an electron revolving in $\mathrm{n}^{\text {th }}$ orbit is given by :
(Symbols have their usual meanings)
The magnetic moment of an electron (e) revolving in an orbit around nucleus with an orbital angular momentum is given by :
A hydrogen atom in ground state absorbs 12.09 eV of energy. The orbital angular momentum of the electron is increased by :
In the following nuclear reaction,
$D\buildrel \alpha \over \longrightarrow {D_1}\buildrel {{\beta ^ - }} \over \longrightarrow {D_2}\buildrel \alpha \over \longrightarrow {D_3}\buildrel \gamma \over \longrightarrow {D_4}$
Mass number of D is 182 and atomic number is 74. Mass number and atomic number of D4 respectively will be _________.
The activity of a radioactive material is 2.56 $\times$ 10$-$3 Ci. If the half life of the material is 5 days, after how many days the activity will become 2 $\times$ 10$-$5 Ci ?
Following statements related to radioactivity are given below :
(A) Radioactivity is a random and spontaneous process and is dependent on physical and chemical conditions.
(B) The number of un-decayed nuclei in the radioactive sample decays exponentially with time.
(C) Slope of the graph of loge (no. of undecayed nuclei) Vs. time represents the reciprocal of mean life time ($\tau$).
(D) Product of decay constant ($\lambda$) and half-life time (T1/2) is not constant.
Choose the most appropriate answer from the options given below :
The Q-value of a nuclear reaction and kinetic energy of the projectile particle, Kp are related as :
Given below are two statements :
Statement I : In hydrogen atom, the frequency of radiation emitted when an electron jumps from lower energy orbit (E1) to higher energy orbit (E2), is given as hf = E1 $-$ E2
Statement II : The jumping of electron from higher energy orbit (E2) to lower energy orbit (E1) is associated with frequency of radiation given as f = (E2 $-$ E1)/h
This condition is Bohr's frequency condition.
In the light of the above statements, choose the correct answer from the options given below :
A hydrogen atom in its ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of :
(Given, Planck's constant = 6.6 $\times$ 10$-$34 Js).
A radioactive nucleus can decay by two different processes. Half-life for the first process is 3.0 hours while it is 4.5 hours for the second process. The effective half-life of the nucleus will be:
How many alpha and beta particles are emitted when Uranium 92U238 decays to lead 82Pb206 ?