When a metallic surface is illuminated with a radiation of wavelength ' λ ', the stopping potential is 'V'. If the same surface is illuminated with radiation of wavelength 6 λ , th

Question

When a metallic surface is illuminated with a radiation of wavelength ' λ ', the stopping potential is 'V'. If the same surface is illuminated with radiation of wavelength 6 λ , the stopping potential is ( 12 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:

$\frac{e V}{12} = \frac{h c}{6 λ} - \frac{h c}{λ _{0}}$

Multiplying the second equation by $12$ gives:

$e V = \frac{2 h c}{λ} - \frac{12 h c}{λ _{0}}$

Equating the two expressions for $e V$ :

$\frac{h c}{λ} - \frac{h c}{λ _{0}} = \frac{2 h c}{λ} - \frac{12 h c}{λ _{0}}$

$\frac{11 h c}{λ _{0}} = \frac{h c}{λ}$

$λ_{0} = 11 λ$

Answer: $11 λ$

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