JEE Main 2022 · 29 Jul · Shift 1 · Physics Q46

← JEE Main 2022 · 29 Jul · Shift 1

PhysicsSingle correct+4 / -1Question 46 of 90

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 :

\dfrac{hc}{E\lambda-hc}

\dfrac{hc\lambda}{E\lambda+hc}

\dfrac{h\lambda}{E\lambda+hc}

\dfrac{hc\lambda}{E\lambda-hc}

Answer and worked solution

Official answer: B

Photoelectric equation. $E=\frac{hc}\lambda-\phi$ and $2E=\frac{hc}{\lambda'}-\phi$ give $\frac{hc}{\lambda'}=E+\frac{hc}\lambda$ , so $\lambda'=\dfrac{hc\lambda}{E\lambda+hc}$ .

(B)

Topics:Photoelectric effect

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JEE Main 2022 · 29 Jul · Shift 1, Physics question 46