Question

A stationary $H{e^ + }$ ion emitted a photon corresponding to a first line of the Lyman series. The photon liberated a photoelectron from a stationary H-atom in the ground state. What is the velocity of a photoelectron?
A. $7.245 \times {10^8}\,cm{s^{ - 1}}$
B. $1.324 \times {10^8}\,cm{s^{ - 1}}$
C. $3.094 \times {10^8}\,cm{s^{ - 1}}$
D. $6.924 \times {10^8}\,cm{s^{ - 1}}$

Answer 1

Hint: In order to solve this question, we will first find the energy of photon emitted by helium ion for Lyman series and this energy will be equal to sum of potential energy of hydrogen atom in ground state and Kinetic energy of the photoelectron, so we will equate both energies and then find the velocity of the photoelectron.

Formula used:
Energy of photon for first line of Lyman series
$E = hc{R_H}{Z^2}[\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}]$
where, $h = 6.625 \times {10^{ - 27}}\,erg\sec $ Planck’s constant, $c = 3 \times {10^{10}}\,cm{s^{ - 1}}$ speed of light, ${R_H} = 109678\,c{m^{ - 1}}$ Rydberg constant and $Z$ denoted atomic number of atom.

Complete answer:
According to the question, we have given that Helium atom release the photon so, atomic number of $H{e^ + }$ is $Z = 2$ so, energy of photon is calculated using the formula
$E = hc{R_H}{Z^2}[\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}]$
On putting the values, we get
$E = 6.625 \times {10^{ - 27}} \times 3 \times {10^{10}} \times 109678 \times 4[\dfrac{3}{4}]$
$\Rightarrow E = 6.54 \times {10^{ - 11}}erg$

Now, this energy will equal the potential energy of the hydrogen atom in the ground state and Kinetic energy of emitted photoelectrons. Potential energy is calculated as $13.6 \times {10^{ - 2}} \times e$ where e is the electronic charge and $13.6 \times {10^{ - 2}}$ is the energy of a hydrogen atom in ground state.
$P.E = 13.6 \times 1.6 \times {10^{ - 21}}erg$
$\Rightarrow P.E = 2.179 \times {10^{ - 11}}erg$

Let $u$ be the velocity of a photoelectron and m is the mass of an electron as $m = 9.108 \times {10^{ - 28}}gm$.
$K.E = \dfrac{1}{2}m{u^2}$
so, equating energy of photon and sum of P.E and K.E we get,
$E = P.E + K.E$
$\Rightarrow 6.54 \times {10^{ - 11}} = 2.179 \times {10^{ - 11}} + \dfrac{1}{2}m{u^2}$
$\Rightarrow (6.54 - 2.179){10^{ - 11}} = \dfrac{1}{2}m{u^2}$
$\Rightarrow {u^2} = \dfrac{{4.361 \times {{10}^{ - 11}} \times 2}}{{9.108 \times {{10}^{ - 28}}}}$
$\therefore u = 3.094 \times {10^8}cm{\sec ^{ - 1}}$

Hence, the correct option is C.

Note: It should be remembered that, all the units of physical quantities measured in CGS system which is Centimetres gram second and also the energy of different levels of hydrogen atom is calculated as $\dfrac{{13.6}}{{{n^2}}}$ where n is the energy level, for ground level $n = 1$ and Lyman series is when particle jumps from first energy level to ground state and it lies in the region of Ultraviolet region.