In aluminum, $4.2 e V$ is required to remove an electron when the light of a wavelength $2000 A^{0}$ falls on this surface. What is its stopping potential?

3 V

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4 V

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2 V

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1 V

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Solution

The correct option is B $2 V$
$ϕ = 4.2 e V$
$λ = 2000 A^{0} = 200 n m$

$e V = \frac{h c}{λ} - ϕ$
$= \frac{1240}{200} - 4.2$
$= (6.2 - 4.2) e V$
Stopping potential = $2 V$

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In aluminum, 4.2eV is required to remove an electron when the light of a wavelength 2000A0 falls on this surface. What is its stopping potential?

The correct option is B 2Vϕ=4.2eVλ=2000A0=200nmeV=hcλ−ϕ =1240200−4.2 =(6.2−4.2)eVStopping potential = 2V

In aluminum, $4.2 e V$ is required to remove an electron when the light of a wavelength $2000 A^{0}$ falls on this surface. What is its stopping potential?

The correct option is B $2 V$
$ϕ = 4.2 e V$
$λ = 2000 A^{0} = 200 n m$

$e V = \frac{h c}{λ} - ϕ$
$= \frac{1240}{200} - 4.2$
$= (6.2 - 4.2) e V$
Stopping potential = $2 V$