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Question: A stationary \(H{e^ + }\) ion emitted a photon corresponding to a first line of the Lyman series. Th...

A stationary He+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. 7.245×108 cms−17.245 \times {10^8}\,cm{s^{ - 1}}
B. 1.324×108 cms−11.324 \times {10^8}\,cm{s^{ - 1}}
C. 3.094×108 cms−13.094 \times {10^8}\,cm{s^{ - 1}}
D. 6.924×108 cms−16.924 \times {10^8}\,cm{s^{ - 1}}

Explanation

Solution

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=hcRHZ2[112−122]E = hc{R_H}{Z^2}[\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}]
where, h=6.625×10−27 ergsec⁡h = 6.625 \times {10^{ - 27}}\,erg\sec Planck’s constant, c=3×1010 cms−1c = 3 \times {10^{10}}\,cm{s^{ - 1}} speed of light, RH=109678 cm−1{R_H} = 109678\,c{m^{ - 1}} Rydberg constant and ZZ denoted atomic number of atom.

Complete answer:
According to the question, we have given that Helium atom release the photon so, atomic number of He+H{e^ + } is Z=2Z = 2 so, energy of photon is calculated using the formula
E=hcRHZ2[112−122]E = hc{R_H}{Z^2}[\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}]
On putting the values, we get
E=6.625×10−27×3×1010×109678×4[34]E = 6.625 \times {10^{ - 27}} \times 3 \times {10^{10}} \times 109678 \times 4[\dfrac{3}{4}]
⇒E=6.54×10−11erg\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 13.6×10−2×e13.6 \times {10^{ - 2}} \times e where e is the electronic charge and 13.6×10−213.6 \times {10^{ - 2}} is the energy of a hydrogen atom in ground state.
P.E=13.6×1.6×10−21ergP.E = 13.6 \times 1.6 \times {10^{ - 21}}erg
⇒P.E=2.179×10−11erg\Rightarrow P.E = 2.179 \times {10^{ - 11}}erg

Let uu be the velocity of a photoelectron and m is the mass of an electron as m=9.108×10−28gmm = 9.108 \times {10^{ - 28}}gm.
K.E=12mu2K.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.EE = P.E + K.E
⇒6.54×10−11=2.179×10−11+12mu2\Rightarrow 6.54 \times {10^{ - 11}} = 2.179 \times {10^{ - 11}} + \dfrac{1}{2}m{u^2}
⇒(6.54−2.179)10−11=12mu2\Rightarrow (6.54 - 2.179){10^{ - 11}} = \dfrac{1}{2}m{u^2}
⇒u2=4.361×10−11×29.108×10−28\Rightarrow {u^2} = \dfrac{{4.361 \times {{10}^{ - 11}} \times 2}}{{9.108 \times {{10}^{ - 28}}}}
∴u=3.094×108cmsec⁡−1\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 13.6n2\dfrac{{13.6}}{{{n^2}}} where n is the energy level, for ground level n=1n = 1 and Lyman series is when particle jumps from first energy level to ground state and it lies in the region of Ultraviolet region.