Question

The de Broglie wavelength associated with an electron accelerated
through a potential difference of 121 V is :​

Answer 1

Given,
x = 121V,


Now,

E = ex
and
also E = hc/λ

So,ex =hc/λ


$λ = \frac{hc}{ex} \\ = \frac{6.626 \times {10}^{ - 34} \times 3 \times {10}^{8} }{1.6 \times {10}^{ - 19} \times 121 } m \\ = 1.02 \times {10}^{ - 8}m$


Symbols hold their usual meaning here except x which is equal to the potential difference i.e Voltage.