Initially a photon of wave length λ1 falls on photocathode and emits an electron of maximum momentum P1. If the wavelength of incident photon is changed to λ2 the maximum momentum of the electron becomes P2. Then value of hc is

P122m=hc1λ1−1λ0&P222m=hc1λ2−1λ0 P12−P222m=hc1λ1−1λ2⇒hc=λ1λ2λ2−λ1×P12−P222m

Initially a photon of wave length λ1 falls on photocathode and emits an electron of maximum momentum P1. If the wavelength of incident photon is changed to λ2 the maximum momentum of the electron becomes P2. Then value of hc is

P122m=hc1λ1−1λ0&P222m=hc1λ2−1λ0 P12−P222m=hc1λ1−1λ2⇒hc=λ1λ2λ2−λ1×P12−P222m

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Initially a photon of wave length $λ_{1}$ falls on photocathode and emits an electron of maximum momentum P₁. If the wavelength of incident photon is changed to $λ_{2}$ the maximum momentum of the electron becomes P₂. Then value of hc is

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$\frac{λ_{1} + λ_{2}}{2 m} (\frac{P_{1}^{2} - P_{2}^{2}}{λ_{1} \cdot λ_{2}})$

$\frac{λ_{1} λ_{2}}{λ_{2} - λ_{1}} (\frac{P_{1}^{2} - P_{2}^{2}}{2 m})$

$\frac{λ_{1} λ_{2}}{λ_{2} - λ_{1}} (\frac{P_{1}^{2}}{P_{2}^{2}})$

$\frac{λ_{1} λ_{2}}{2 m} (\frac{P_{1}^{2} + P_{2}^{2}}{λ_{2} - λ_{1}})$

answer is B.

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Detailed Solution

$\frac{P_{1}^{2}}{2 m} = hc (\frac{1}{λ_{1}} - \frac{1}{λ_{0}}) & \frac{P_{2}^{2}}{2 m} = hc (\frac{1}{λ_{2}} - \frac{1}{λ_{0}}) \frac{P_{1}^{2} - P_{2}^{2}}{2 m} = hc (\frac{1}{λ_{1}} - \frac{1}{λ_{2}}) \Rightarrow hc = \frac{λ_{1} λ_{2}}{λ_{2} - λ_{1}} \times (\frac{P_{1}^{2} - P_{2}^{2}}{2 m})$