Work function of nickel is 5eV. When a light of wavelength 2000Å falls on it, it emits photoelectrons in the circuit. Then the potential difference (in volt) necessary to stop the fastest electrons emitted is h=6.4×10−34Js

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Work function of nickel is $5 e V$ . When a light of wavelength $2000 Å$ falls on it, it emits photoelectrons in the circuit. Then the potential difference (in volt) necessary to stop the fastest electrons emitted is $(h = 6.4 \times 10^{- 34} J s)$

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Detailed Solution

$E = ϕ + eV \Rightarrow \frac{hc}{λ} = ϕ + eV \Rightarrow \frac{6.4 \times 10^{- 34} \times 3 \times 10^{8}}{2000 \times 10^{- 10} \times 1.6 \times 10^{- 19}} - 5 e = e V$