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(A) According to Einstein's photoelectric equation: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_{0}} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda

Vedclass · Accepted Answer

$4 \lambda$

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(A) According to Einstein's photoelectric equation: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_{0}} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)$.For the first case: $3 eV_{s} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)$  --- $(i)$For the second case: $eV_{s} = hc \left( \frac{1}{2 \lambda} - \frac{1}{\lambda_{0}} \right)$  --- (ii)Dividing equation $(i)$ by equation (ii):$\frac{3 eV_{s}}{eV_{s}} = \frac{hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)}{hc \left( \frac{1}{2 \lambda} - \frac{1}{\lambda_{0}} \right)}$$3 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_{0}}}{\frac{1}{2 \lambda} - \frac{1}{\lambda_{0}}}$$3 \left( \frac{1}{2 \lambda} - \frac{1}{\lambda_{0}} \right) = \frac{1}{\lambda} - \frac{1}{\lambda_{0}}$$\frac{3}{2 \lambda} - \frac{3}{\lambda_{0}} = \frac{1}{\lambda} - \frac{1}{\lambda_{0}}$$\frac{3}{2 \lambda} - \frac{1}{\lambda} = \frac{3}{\lambda_{0}} - \frac{1}{\lambda_{0}}$$\frac{1}{2 \lambda} = \frac{2}{\lambda_{0}}$$\lambda_{0} = 4 \lambda$.

Similar Questions

When a metallic surface is illuminated with light of wavelength $\lambda$,the stopping potential is $x \ V$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential is $\frac{x}{3} \ V$. The threshold wavelength for the metallic surface is ..........

If the stopping potential of photoelectrons emitted by a source placed at a distance of $0.2 \ m$ from a photocell is $0.6 \ V$,then the stopping potential of photoelectrons emitted by the same source placed at a distance of $0.6 \ m$ from the photocell will be .......... $V$.

The maximum velocity of an electron emitted by light of wavelength $\lambda$ incident on the surface of a metal of work function $\phi$ is:
Where $h =$ Planck's constant, $m =$ mass of electron, and $c =$ speed of light.

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