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Question

If $$\lambda_0$$ and $$\lambda$$ be the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is:


A
2hm(λ0λ)
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B
2hcm(λ0λ)
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C
2hcm(λ0λλλ0)
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D
2hm(1λ01λ)
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Solution

The correct option is C $$\displaystyle \sqrt{\frac{2hc}{m} \left ( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right )}$$
If $$\lambda_0$$ and $$\lambda$$ be the threshold wavelength and wavelength of incident light then according to the equation
$$\dfrac{hc}{\lambda}=\dfrac{hc}{\lambda_0} + \dfrac{1}{2}mv^2$$
$$v=\sqrt{\dfrac{2hc}{m}(\dfrac{1}{\lambda}-\dfrac{1}{\lambda_0})}$$
$$v=\sqrt{\dfrac{2hc}{m}(\dfrac{\lambda_0 - \lambda}{\lambda \lambda_0})}$$

Chemistry

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