Verified Solution • 2025–26 Session moderate

QJEEM22Ph2640: JEE Main 2022 Physics - Class 12 JEEM22Ph2640 Solution: The kinetic energy of emitted electron is E when the light…

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

Multiple Choice Options

  • A hcEλ−hc\frac{\mathrm{hc}}{\mathrm{E} \lambda-\mathrm{hc}}
  • B hcλEλ+hc\frac{\mathrm{hc} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}
  • C hλEλ+hc\frac{\mathrm{h} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}
  • D  hc λEλ−hc\frac{\text { hc } \lambda}{\mathrm{E} \lambda-\mathrm{hc}}

Step-by-Step Verified Solution

k=hcλ−ϕ=Ek = {{hc} \over \lambda } - \phi = E

and, 2k=hcλ2−ϕ=2E2k = {{hc} \over {{\lambda _2}}} - \phi = 2E

⇒hcλ−E=hcλ2−2E\Rightarrow {{hc} \over \lambda } - E = {{hc} \over {{\lambda _2}}} - 2E

⇒hcλ2=hcλ+E\Rightarrow {{hc} \over {{\lambda _2}}} = {{hc} \over \lambda } + E

⇒λ2=hcλhc+λE\Rightarrow {\lambda _2} = {{hc\lambda } \over {hc + \lambda E}}

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