The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 A is 0.42 V. If the threshold frequency is x × 1013 /s, where x is _________ (nearest integer). (Given, speed light = 3 × 108 m/s, Planck's constant = 6.63 × 10 - 34 Js) Numerical answer type. Work it out, then open the answer.

Answer: 35. E = (hc)/(λ) = 6.63 × 10⁻³⁴ × 3 × 10⁸6630 × 10⁻¹⁰ = 3.0 × 10⁻¹⁹ J φ = E - eV₀ = 3.0 × 10⁻¹⁹ - 0.42 × 1.6 × 10⁻¹⁹ = 2.33 × 10⁻¹⁹ J ν₀ = φ/h = 2.33 × 10⁻¹⁹6.63 × 10⁻³⁴ = 3.51 × 10¹⁴ = 35.1 × 10¹³ Hz So x = 35.

The stopping potential for photoelectrons emitted from a surface… | JEE Main 2022 (26 Jun, Shift 2)

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Photoelectric Effect: Einstein's Equation

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 $\mathring{A}$ is 0.42 V. If the threshold frequency is x $\times$ 1013 /s, where x is _________ (nearest integer). (Given, speed light = 3 $\times$ 108 m/s, Planck's constant = 6.63 $\times$ 10 $-$ 34 Js)

Numerical answer type. Work it out, then open the answer.

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Answer: 35

$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{6630 \times 10^{-10}} = 3.0 \times 10^{-19}$ J
$\phi = E - eV_0 = 3.0 \times 10^{-19} - 0.42 \times 1.6 \times 10^{-19} = 2.33 \times 10^{-19}$ J
$\nu_0 = \frac{\phi}{h} = \frac{2.33 \times 10^{-19}}{6.63 \times 10^{-34}} = 3.51 \times 10^{14} = 35.1 \times 10^{13}$ Hz
So $x = 35$ .

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