A filter transmits only the radition of wavelength greater than 440 nm. Radition from a hydrogen discharge tube goes through such a filter and is incident on a metal of work function 2.0 e V. Find the stopping potential which can stop the photoelectrons.

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Solution

$λ$ = 440 nm,\\

$Θ$ = Charge of the electron

$Φ$ = 2 e V,

$v_{0}$ = Stopping potential

We have $\frac{h c}{λ} - Φ = e V_{0}$

$\Rightarrow \frac{4.14 \times 10^{- 15} \times 3 \times 10^{8}}{440 \times 10^{- 9}} - 2 e V = e V_{0} e V_{0} = (\frac{1242}{440} - 2) e V = 0.823 e V .$

$\Rightarrow v_{0} = 0.823 V o l t s$

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A filter transmits only the radition of wavelength greater than 440 nm. Radition from a hydrogen discharge tube goes through such a filter and is incident on a metal of work function 2.0 e V. Find the stopping potential which can stop the photoelectrons.

λ = 440 nm,\\ Θ = Charge of the electron Φ = 2 e V, v0 = Stopping potential We have hcλ−Φ=eV0 ⇒4.14×10−15×3×108440×10−9−2eV=eV0eV0=(1242440−2)eV=0.823eV. ⇒v0=0.823Volts