Question

Ultraviolet light of wavelength ${\lambda }_{1}and{\lambda }_2({\lambda }_2>{\lambda }_1)$ when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energies $E_{1}and{E}_2$ respectively. The value of planck's constant can be found from the relation

• A. $h=\frac{1}{c}({\lambda }_2-{\lambda }_1)(E_1-E_2)$
• B. $h=\frac{1}{c}({\lambda }_1+{\lambda }_2)(E_1+E_2)$
• C. $h=\frac{(E_1-E_2){\lambda }_1{\lambda }_2}{c({\lambda }_2-{\lambda }_1)}$
• D. $h=\frac{(E_1+E_2){\lambda }_1{\lambda }_2}{c({\lambda }_1+{\lambda }_2)}$

Answer 1

The correct option is C $h=\frac{(E_1-E_2){\lambda }_1{\lambda }_2}{c({\lambda }_2-{\lambda }_1)}$

Using Einstein's formula, $E_1=\frac{hC}{{\lambda }_1}-13.6eV;E_2=\frac{hC}{{\lambda }_2}-13.6eV$
Or, $E_1-E_2=hC[\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}];E_1-E_2=hC\left[\frac{{\lambda }_2-{\lambda }_1}{{\lambda }_1{\lambda }_2}\right]$

Or, $h=\frac{(E_1-E_2){\lambda }_1{\lambda }_2}{c({\lambda }_2-{\lambda }_1)}$