Question

The work function for the surface of aluminium is 4.2 eV. How much potential difference will be required to stop the emission of maximum energy electron emitted by light of 2000 $\overset{\circ}{A}$ wavelength ? What will be the wavelength of that incident light for which stopping potential will be zero ?
(Ans: 2.016 V, 2959 $\overset{\circ}{A}$)

Answer 1

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Answer 2

Hi Students,

● Answer -

PD = 2.006 eV

λ = 2955 A°

● Explaination -

Energy threshold for 2000 A° wavelength is calculated by -

E = hc / λ

E = (6.62×10^-34 × 3×10^8) / (2000×10^-10)

E = 9.93×10^-19 J

E = 6.206 eV

Therefore, required potential difference is given by -

PD = E - V

PD = 6.206 - 4.2

PD = 2.006 eV

Wavelength of incident light with stopping potential zero can be calculated by -

λ = hc / V

λ = (6.62×10^-34 × 3×10^8) / (4.2×1.6×10^-19)

λ = 2.955×10^-7 m

λ = 2955 A°

Hope this helps..