Atoms and Nuclei
The speed of the electron in a hydrogen atom in the $n=3$ level is
(Planck's constant $=6.6 \times 10^{-34} \mathrm{Js}$ )
One mole of radium has an activity of $\frac{1}{3.7}$ kilo curie. Its decay constant is
(Avagadro number $=6 \times 10^{23} \mathrm{~mol}^{-1}$ )
Given below are two statements: one is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$
Assertion A : The binding energy per nucleon is practically independent of the atomic number for nuclei of mass number in the range 30 to 170 .
Reason R : Nuclear force is short ranged.
In the light of the above statements, choose the correct answer from the options given below
$_{92}^{238}A \to _{90}^{234}B + _2^4D + Q$
In the given nuclear reaction, the approximate amount of energy released will be:
[Given, mass of ${ }_{92}^{238} \mathrm{~A}=238.05079 \times 931.5 ~\mathrm{MeV} / \mathrm{c}^{2},$
mass of ${ }_{90}^{234} B=234 \cdot 04363 \times 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2},$
mass of $\left.{ }_{2}^{4} D=4 \cdot 00260 \times 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2}\right]$
A $12.5 \mathrm{~eV}$ electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:
The energy of $\mathrm{He}^{+}$ ion in its first excited state is, (The ground state energy for the Hydrogen atom is $-13.6 ~\mathrm{eV})$ :
Two radioactive elements A and B initially have same number of atoms. The half life of A is same as the average life of B. If $\lambda_{A}$ and $\lambda_{B}$ are decay constants of A and B respectively, then choose the correct relation from the given options.
The half life of a radioactive substance is T. The time taken, for disintegrating $\frac{7}{8}$th part of its original mass will be:
The angular momentum for the electron in Bohr's orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be
A radio active material is reduced to $1 / 8$ of its original amount in 3 days. If $8 \times 10^{-3} \mathrm{~kg}$ of the material is left after 5 days the initial amount of the material is
The waves emitted when a metal target is bombarded with high energy electrons are
For a nucleus ${ }_{\mathrm{A}}^{\mathrm{A}} \mathrm{X}$ having mass number $\mathrm{A}$ and atomic number $\mathrm{Z}$
A. The surface energy per nucleon $\left(b_{\mathrm{s}}\right)=-a_{1} A^{2 / 3}$.
B. The Coulomb contribution to the binding energy $\mathrm{b}_{\mathrm{c}}=-a_{2} \frac{Z(Z-1)}{A^{4 / 3}}$
C. The volume energy $\mathrm{b}_{\mathrm{v}}=a_{3} A$
D. Decrease in the binding energy is proportional to surface area.
E. While estimating the surface energy, it is assumed that each nucleon interacts with 12 nucleons. ( $a_{1}, a_{2}$ and $a_{3}$ are constants)
Choose the most appropriate answer from the options given below:
A small particle of mass $m$ moves in such a way that its potential energy $U=\frac{1}{2} m ~\omega^{2} r^{2}$ where $\omega$ is constant and $r$ is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of $n^{\text {th }}$ orbit will be proportional to,
The energy levels of an hydrogen atom are shown below. The transition corresponding to emission of shortest wavelength is :
An electron of a hydrogen like atom, having $Z=4$, jumps from $4^{\text {th }}$ energy state to $2^{\text {nd }}$ energy state. The energy released in this process, will be :
(Given Rch = $13.6~\mathrm{eV}$)
Where R = Rydberg constant
c = Speed of light in vacuum
h = Planck's constant
The mass of proton, neutron and helium nucleus are respectively $1.0073~u,1.0087~u$ and $4.0015~u$. The binding energy of helium nucleus is :
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} $