A monochromatic light of intensity 8 mW emits $5 \times 10^{15}$ photons per second. The light ejects photoelectrons from a metal surface. The stopping potential for this set up is 3V. What is the work function of the metal?

3 eV

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7 eV

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10 eV

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5 eV

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Solution

The correct option is B 7 eV
$\frac{h c}{λ} = 10 e V$
$e V_{0} = \frac{h c}{λ} - ϕ$
$ϕ = \frac{h c}{λ} - e V_{0}$
=10-3
=7 eV

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A monochromatic light of intensity 8 mW emits 5×1015 photons per second. The light ejects photoelectrons from a metal surface. The stopping potential for this set up is 3V. What is the work function of the metal?

The correct option is B 7 eVhcλ=10eV eV0=hcλ−ϕ ϕ=hcλ−eV0 =10-3 =7 eV

A monochromatic light of intensity 8 mW emits $5 \times 10^{15}$ photons per second. The light ejects photoelectrons from a metal surface. The stopping potential for this set up is 3V. What is the work function of the metal?

The correct option is B 7 eV
$\frac{h c}{λ} = 10 e V$
$e V_{0} = \frac{h c}{λ} - ϕ$
$ϕ = \frac{h c}{λ} - e V_{0}$
=10-3
=7 eV