The kinetic energy of emitted electron is E when the light incident on the metal has wavelength λ. To double the kinetic energy, the incident light must have wavelength: (a) (hc)/(Eλ - hc) (b) (hcλ)/(Eλ + hc) (c) (hλ)/(Eλ + hc) (d) (hcλ)/(Eλ - hc)

Answer: (b). E = (hc)/(λ) - φ and 2E = (hc)/(λ') - φ. Subtract: (hc)/(λ') = E + (hc)/(λ) = (Eλ + hc)/(λ) So the answer is (b) λ' = (hcλ)/(Eλ + hc).

The kinetic energy of emitted electron is E when the light incident… | JEE Main 2022 (29 Jul, Shift 1)

Q97JEE Main 2022 · 29 JUL · SHIFT 1Medium

Photoelectric Effect: Einstein's Equation

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

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Answer: (b)

$E = \frac{hc}{\lambda} - \phi$ and $2E = \frac{hc}{\lambda'} - \phi$ .
Subtract: $\frac{hc}{\lambda'} = E + \frac{hc}{\lambda} = \frac{E\lambda + hc}{\lambda}$
So the answer is (b) $\lambda' = \frac{hc\lambda}{E\lambda + hc}$ .

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