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In a photoelectric emission experiment, the stopping potential for a given metal is $V$ volt, when radiation of wavelength $\lambda$ is used. If radiation of wavelength $2 \lambda$ is used with the same metal, then the stopping potential (in volt) will be
$\text { [ } c \text { = velocity of light, } e=\text { charge on electron, } h=\text { Planck's constant }]$
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$\text { [ } c \text { = velocity of light, } e=\text { charge on electron, } h=\text { Planck's constant }]$
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Verified Answer
The correct answer is:
$V-\frac{h c}{2 e \lambda}$
According to Einstein's equation of photo-electric effect:
$e V=\frac{h c}{\lambda}-\Phi---(1)$
where, stopping potential is $V, \lambda$ the incident wavelength and $\Phi$ the work-function of the metal.
Now, if the incident wavelength is switched to $2 \lambda$, the stopping potential can be written as
$e V^{\prime}=\frac{h c}{2 \lambda}-\Phi---(2)$
On subtracting equation (1) and (2), and subsequently rearranging:
$V^{\prime}=V-\frac{h c}{2 \lambda e}$
$e V=\frac{h c}{\lambda}-\Phi---(1)$
where, stopping potential is $V, \lambda$ the incident wavelength and $\Phi$ the work-function of the metal.
Now, if the incident wavelength is switched to $2 \lambda$, the stopping potential can be written as
$e V^{\prime}=\frac{h c}{2 \lambda}-\Phi---(2)$
On subtracting equation (1) and (2), and subsequently rearranging:
$V^{\prime}=V-\frac{h c}{2 \lambda e}$
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