Question

Absorption of energy by an atom of hydrogen in the ground state results in the ejection of electron with de-Broglie wavelength $\lambda =4.7\times {10}^{-10}$ m. Given that the ionization energy is 13.6eV, calculate the energy of photon which caused the ejection of electron.

• A. $4.271\times {10}^{-18}J$
• B. $3.271\times {10}^{-18}J$
• C. $3.271\times {10}^{-16}J$
• D. $4.271\times {10}^{-16}J$

Answer 1

The correct option is B $3.271\times {10}^{-18}J$

The energy of photon which can cause ejection of electron = kinetic energy of ejected electron + ionization energy ...(i)
Ionization energy $=13.6eV$
$=\frac{13.6}{6.24\times {10}^{18}}J=2.179\times {10}^{-18}J$ ...(ii)
Here $6.24\times {10}^{18}eV=1J;$ de-broglie wavelength $=\lambda =\frac{h}{mv}$
$\therefore v=\frac{h}{m\lambda }=\frac{6.626\times {10}^{-34}}{9.1\times {10}^{-31}\times 4.7\times {10}^{-10}}=0.15492\times {10}^7$ m/sec
K.E.$=\frac{1}{2}m{v}^2=\frac{1}{2}\times 9.1\times {10}^{-31}\times (0.15492\times {10}^7{)}^2=1.092\times {10}^{-18}J$ ...(iii)
Now from equation (i)
The energy of photon which can cause ejection of electron
$=2.179\times {10}^{-18}+1.092\times {10}^{-18}=3.271\times {10}^{-18}J$