Question

Energy of a photon whose de-Broglie wavelength is equal to the wavelength of an electron accelerated through a potential difference of $125V$ is near to

• A. 11.5 eV
• B. 11.5 keV
• C. 125 eV
• D. 1250 eV

Answer 1

Answer: (C)

Given that Potential difference $V=125V$
Then, $\lambda =\frac{12375}{V}\mathring{A}$
$\lambda =\frac{12375}{125}\mathring{A}$
$\therefore$ Energy of electron in $eV$
$E=\frac{hc}{\lambda }$
$E=\frac{6.6\times 10^{-34}\times 3\times 10^8\times 125\times 10^{-10}}{12375}=125eV$
Then the energy of photon $=125eV$