Dual Nature of Matter

JEE Main level, single correct, medium

37 of 200

Question 37 of 200

The kinetic energy of emitted electron is EE when the light incident on the metal has wavelength λ\lambda. To double the kinetic energy, the incident light must have wavelength:

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Answer: (B) hcλEλ+hc\dfrac{hc\lambda }{E\lambda +hc}

Using Einstein's photoelectric equation, kinetic energy of an electron is EE when the incident wavelength is λ\lambda is E=hcλ−ϕ...(1)E=\dfrac{hc}{\lambda }-\phi ...\left(1\right)
When kinetic energy doubles, then, 2E=hcλ′−ϕ...(2)2E=\dfrac{hc}{{\lambda }^{'}}-\phi ...\left(2\right)
Subtracting (2)−(1)\left(2\right)-\left(1\right), we have
E=hc(1λ′−1λ)E=hc\left(\dfrac{1}{{\lambda }^{'}}-\dfrac{1}{\lambda }\right)
⇒λ′=hcλEλ+hc\Rightarrow {\lambda }^{'}=\dfrac{hc\lambda }{E\lambda +hc}

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