When a certain metal surface is illuminated with a light of wavelength λ , the stopping potential is V . When the same surface is illuminated by light of wavelength 2 λ , the stopping potential is ( 3

Question

When a certain metal surface is illuminated with a light of wavelength λ , the stopping potential is V . When the same surface is illuminated by light of wavelength 2 λ , the stopping potential is ( 3 V ​ ) . The 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 first case:

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

For the second case:

$\frac{e V}{3} = \frac{h c}{2 λ} - \frac{h c}{λ _{0}}$

Multiplying the second equation by $3$ gives:

$e V = \frac{3 h c}{2 λ} - \frac{3 h c}{λ _{0}}$

Equating the two expressions for $e V$ :

$\frac{h c}{λ} - \frac{h c}{λ _{0}} = \frac{3 h c}{2 λ} - \frac{3 h c}{λ _{0}}$

Dividing by $h c$ and rearranging the terms:

$\frac{3}{λ _{0}} - \frac{1}{λ _{0}} = \frac{3}{2 λ} - \frac{1}{λ}$

$\frac{2}{λ _{0}} = \frac{1}{2 λ}$

$λ_{0} = 4 λ$

Answer: $4 λ$

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