In a photoelectric effect measurement, the stopping potential for a given metal is found to be V 0 volts when radiation of wavelength λ 0 is used. If radiation of wavelength 2 λ 0 is used with t

Question

In a photoelectric effect measurement, the stopping potential for a given metal is found to be $V_{0}$ volts when radiation of wavelength $λ_{0}$ is used. If radiation of wavelength $2 λ_{0}$ is used with the same metal, then the stopping potential will be observed to be (Where symbols have their usual meaning):

TestHub · Accepted Answer

The photoelectric equation is given by:

$e V_{0} = \frac{h c}{λ} - ϕ$ ...(1)

where $V_{0}$ is the stopping potential, $λ$ is the wavelength of the incident light, $h$ is Planck's constant, $c$ is the speed of light, $e$ is the elementary charge, and $ϕ$ is the work function of the metal.

In the first case, we have:

$e V_{0} = \frac{h c}{λ _{0}} - ϕ$ ...(1)

In the second case, the wavelength is $2 λ_{0}$ , so the stopping potential $V_{0}^{'}$ is:

$e V_{0}^{'} = \frac{h c}{2 λ _{0}} - ϕ$ ...(2)

We want to find $V_{0}^{'}$ in terms of $V_{0}$ . We can rearrange equation (1) to express the work function $ϕ$ :

$ϕ = \frac{h c}{λ _{0}} - e V_{0}$

Substitute this expression for $ϕ$ into equation (2):

$e V_{0}^{'} = \frac{h c}{2 λ _{0}} - (\frac{h c}{λ _{0}} - e V_{0})$

$e V_{0}^{'} = \frac{h c}{2 λ _{0}} - \frac{h c}{λ _{0}} + e V_{0}$

$e V_{0}^{'} = e V_{0} - \frac{h c}{2 λ _{0}}$

$V_{0}^{'} = V_{0} - \frac{h c}{2 e λ _{0}}$

Therefore, the stopping potential in the second case is $V_{0} - \frac{h c}{2 e λ _{0}}$ .

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