AP EAPCET 23-Aug-2021 Shift 1 Paper

Given, wavelength of electron, λ=0.5Å
By using de-Broglie wavelength,λ=

h

√2meV

where, h is Planck's constant =6.63×10−34J−S,
m is mass of electron,
e is charge of electron
and V is applied voltage.
∴V=

h2

2meλ2

=

(6.63×10−34)2

2×91×10−31×1.6×10−19×(0.5×10−10)2

=602V