Q.a) Under what potential difference should an electron be accelerated to obtain de Broglie wavelength of 0.6 Å? (h = 6.62×10^-34 J-s, m_e = 9.1×10^-31 kg) b) Give an example of production of electron by photon. OR

a) Using λ = h/√(2meV), the accelerating potential needed for λ = 0.6 Å works out to about 418 V. b) The photoelectric effect (a photon ejecting an electron from a metal surface) is an example of electron production by a photon.

a) Accelerating potential for given de Broglie wavelength:

For an electron accelerated through potential difference $V$, its kinetic energy is $eV = \dfrac{p^2}{2m}$, so momentum $p=\sqrt{2meV}$. The de Broglie wavelength is

$\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2meV}} \implies V = \dfrac{h^2}{2me\lambda^2}$

Substituting $h=6.62\times10^{-34}$ J·s, $m=9.1\times10^{-31}$ kg, $e=1.6\times10^{-19}$ C, $\lambda=0.6\times10^{-10}$ m:

$h^2 = 4.382\times10^{-67}, \qquad 2me\lambda^2 = 2(9.1\times10^{-31})(1.6\times10^{-19})(0.36\times10^{-20}) = 1.048\times10^{-69}$

$V = \dfrac{4.382\times10^{-67}}{1.048\times10^{-69}} \approx 418\ \text{V}$

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