AP EAPCET 20-Aug-2021 Shift 2 Paper

As we know that,
By using de-Broglie wavelength,
λ=

h

p

=

h

√2‌mE

=

h

√2‌meV

where, λ is wavelength of X-ray,
h is Planck's constant i.e. 6.63×10−34J−s, p is momentum, m is mass, E is energy, e is charge and V is electric potential.
∴λ∝

1

√V