Light is incident on the cathode of a photocell and the stopping voltages are measured for light of two different wavelengths. From the data given below, calculate the value of the universal constant $\frac{h c}{e}$ and the work function of the metal of the cathode in $eV$ .
Wavelength (Å) Stopping voltage (Volt)
4000 1.3
4500 0.9

Wavelength (Å)	Stopping voltage (Volt)
4000	1.3
4500	0.9

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$8 \times 10^{- 7} Vm, 1.2 eV$

$1.44 \times 10^{- 6} Vm, 2.3 eV$

$3 \times 10^{- 6} Vm, 5 eV$

$2 \times 10^{- 6} Vm, 3.2 eV$

answer is C.

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Detailed Solution

From photo-electricity equation

$\frac{h c}{λ_{1}} = e {V_{s}}_{1} + ϕ_{0}$ and $\frac{h c}{λ_{2}} = e {V_{s}}_{2} + ϕ_{0}$

From these two equations, we have

$\frac{h c}{e} = \frac{({V_{s}}_{2} - {V_{s}}_{1}) (λ_{1} λ_{2})}{(λ_{1} - λ_{2})}$ $= \frac{(1.3 - 0.9) [(4000 \times 10^{- 10}) \times (4500 \times 10^{- 10})]}{500 \times 10^{- 10}}$ $= 1.44 \times 10^{- 6} Vm$

Also we have, ${V_{s}}_{1} = \frac{h c}{e λ_{1}} - \frac{ϕ_{0}}{e}$

$\Rightarrow \frac{ϕ_{0}}{e} = \frac{h c}{e λ_{1}} - {V_{s}}_{1} = \frac{1.44 \times 10^{- 6}}{4000 \times 10^{- 10}} - 1.3$

$\Rightarrow ϕ_{0} = 2.3 eV$

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