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Standard XII

Physics

Stopping Potential Revisited

In a photoele...

Question

In a photoelectric experiment the stopping potential for the incident light of wavelength $4000 ˚ A$ is 2 volt. If the wavelength be changed to $3000 ˚ A$ , the stopping potential will be :

2 V

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Zero

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Less than 2 V

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More than 2 V

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Solution

The correct option is D More than 2 V
Photoelectric Effect,

$K E_{m a x} = h v - W_{o}$

The stopping potential,

$V_{s} e = K E_{m a x}$

$V_{s} e = h v - W_{o}$

$V_{s} e = \frac{h c}{λ} - W_{o}$

From the above equation, as $λ$ decreases, $V_{s}$ increases.
Hence $V_{s} > 2 V$

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In a photoelectric experiment the stopping potential for the incident light of wavelength 4000˚A is 2 volt. If the wavelength be changed to 3000˚A, the stopping potential will be :

The correct option is D More than 2 VPhotoelectric Effect,KEmax=hv−WoThe stopping potential,Vse=KEmaxVse=hv−WoVse=hcλ−WoFrom the above equation, as λ decreases, Vs increases.Hence Vs>2V

In a photoelectric experiment the stopping potential for the incident light of wavelength $4000 ˚ A$ is 2 volt. If the wavelength be changed to $3000 ˚ A$ , the stopping potential will be :

The correct option is D More than 2 V
Photoelectric Effect,

$K E_{m a x} = h v - W_{o}$

The stopping potential,

$V_{s} e = K E_{m a x}$

$V_{s} e = h v - W_{o}$

$V_{s} e = \frac{h c}{λ} - W_{o}$

From the above equation, as $λ$ decreases, $V_{s}$ increases.
Hence $V_{s} > 2 V$