Question

Energy of a photon whose de-Broglie wavelength is equal to the wavelength of an electron accelerated through a potential difference of $125 \,V$ 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 $V =125 \,V$
Then, $\lambda=\frac{12375}{V} \mathring{A}$
$\lambda=\frac{12375}{125} \mathring{A}$
$ \therefore $ Energy of electron in $eV$
$E=\frac{h c}{\lambda}$
$E=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8} \times 125 \times 10^{-10}}{12375}=125\, eV$
Then the energy of photon $=125\, eV$