Photoelectric emission takes place from a cer | vedclass

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Question

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by $K_{max} = E - \Phi$

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$\sqrt{\frac{6hv}{m}}$

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by $K_{max} = E - \Phi$,where $E$ is the energy of the incident photon and $\Phi$ is the work function of the metal.Given the threshold frequency is $v$,the work function is $\Phi = hv$.The energy of the incident radiation with frequency $4v$ is $E = h(4v) = 4hv$.Substituting these values into the equation:$K_{max} = 4hv - hv = 3hv$.Since $K_{max} = \frac{1}{2}mv_{max}^2$,we have:$\frac{1}{2}mv_{max}^2 = 3hv$.Solving for $v_{max}$:$v_{max}^2 = \frac{6hv}{m}$.$v_{max} = \sqrt{\frac{6hv}{m}}$.

Photoelectric emission takes place from a certain metal at threshold frequency $v$. If the radiation of frequency $4v$ is incident on the metal plate,the maximum velocity of the emitted photoelectrons will be ($m=$ mass of photoelectron,$h=$ Planck's constant).

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