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. $h=\frac{1}{c}({\lambda }_2+{\lambda }_1)(E_1+E_2)$
• 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

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

Suppose the frequency of the photons are ${\nu }_1$ and ${\nu }_2$ (${\nu }_1$>${\nu }_1$). Where, $\nu =c/\lambda$.
According to the question:
$h{\nu }_1-E_0=E_1$
$h{\nu }_2-E_0=E_2$ , where, $E_0$ is the ionization energy of the hydrogen.
Solving this two equation we will get:

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

So, the answer is option (C).