The de Broglie wavelength (lambda) associated | vedclass

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(D) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$,where $v$ is the velocity of the photoelectron.According to Einstein's photoelectri

Vedclass · Accepted Answer

$\lambda \propto \frac{1}{(v - v_0)^{\frac{1}{2}}}$

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(D) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$,where $v$ is the velocity of the photoelectron.According to Einstein's photoelectric equation:$hv = hv_0 + KE$$hv - hv_0 = \frac{1}{2}mv^2$$2h(v - v_0) = mv^2$From this,the velocity $v$ is proportional to $(v - v_0)^{\frac{1}{2}}$:$v = \sqrt{\frac{2h(v - v_0)}{m}} \propto (v - v_0)^{\frac{1}{2}}$Substituting this into the de Broglie equation:$\lambda = \frac{h}{mv} \propto \frac{1}{(v - v_0)^{\frac{1}{2}}}$Thus,$\lambda \propto \frac{1}{(v - v_0)^{\frac{1}{2}}}$.

The de Broglie wavelength $(\lambda)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as,[$v_0$ is threshold frequency]:

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