When a metallic surface is illuminated with radiation of wavelength λ , the stopping potential is V . If the surface is now illuminated with a radiation of wavelength 4 λ , the stopping potential ( 8

Question

When a metallic surface is illuminated with radiation of wavelength λ , the stopping potential is V . If the surface is now illuminated with a radiation of wavelength 4 λ , the stopping potential ( 8 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:

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

Multiplying the second equation by $8$ gives:

$e V = \frac{2 h c}{λ} - \frac{8 h c}{λ _{0}}$

Equating the two expressions for $e V$ :

$\frac{h c}{λ} - \frac{h c}{λ _{0}} = \frac{2 h c}{λ} - \frac{8 h c}{λ _{0}}$

Rearranging the terms:

$\frac{8 h c}{λ _{0}} - \frac{h c}{λ _{0}} = \frac{2 h c}{λ} - \frac{h c}{λ}$

$\frac{7 h c}{λ _{0}} = \frac{h c}{λ}$

$λ_{0} = 7 λ$

Answer: $7 λ$

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