When light of wavelength ' λ ' is incident on a photosensitive surface, the stopping potential is 'V'. When a light of wavelength 1.5 λ is incident on the same surface, the stoppin

Question

When light of wavelength ' λ ' is incident on a photosensitive surface, the stopping potential is 'V'. When a light of wavelength 1.5 λ is incident on the same surface, the stopping potential is ' 4 V ​ '. Threshold wavelength for the surface is

TestHub · Accepted Answer

Using Einstein's photoelectric equation:

$e V = \frac{h c}{λ} - \frac{h c}{λ _{0}}$

For the second case:

$e (\frac{V}{4}) = \frac{h c}{1.5 λ} - \frac{h c}{λ _{0}}$

Multiplying the second equation by $4$ gives:

$e V = \frac{4 h c}{1.5 λ} - \frac{4 h c}{λ _{0}} = \frac{8 h c}{3 λ} - \frac{4 h c}{λ _{0}}$

Equating the two expressions for $e V$ :

$\frac{h c}{λ} - \frac{h c}{λ _{0}} = \frac{8 h c}{3 λ} - \frac{4 h c}{λ _{0}}$

Dividing by $h c$ and rearranging the terms:

$\frac{4}{λ _{0}} - \frac{1}{λ _{0}} = \frac{8}{3 λ} - \frac{1}{λ}$

$\frac{3}{λ _{0}} = \frac{5}{3 λ}$

$λ_{0} = \frac{9}{5} λ$

Answer: $\frac{9}{5} λ$

Physics - Modern Physics Question with Solution | TestHub

Options:

Doubts & Discussion