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, it is found to liberate electrons with kinetic energies $E_1$ and $E_2$ respectively. The value of Planck's constant can be found from which of the relation

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

Answer 1

Answer: (C)

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

If $W_0$ is the ionization energy of hydrogen atom, the incident energy first ionizes the atom and rest of the energy is given in the kinetic energy.

$\frac{hc}{{\lambda }_1}=W_0+E_1$

$\frac{hc}{{\lambda }_2}=W_0+E_2$

Subtracting the two equations gives,

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

So, the answer is option (C).