Question

Calculate the
A) momentum , and
B) de broglie wavelength of the electrons accelerated through a potential difference of 56 v.
Plzzzzzzzz help me to solve this.........

Answer 1

Given,

Potential difference, V = 56 V

Energy of electron accelerated,

$= 56 \: ev = 56 \times 1.6 \times 10 {}^{ - 19}j$

(a) As, Energy, E =

$\frac{p {}^{2} }{2m}$

[ p = mv, E = ½ mv² ]

∴ p² = 2mE

⇒ p = √2mE

⇒

$p = \sqrt{2 \times 9 \times 10 {}^{ - 31} \times 56 \times 1.6 \times 10 {}^{ - 19} }$

$p = 4.02 \times 10 {}^{ - 24} kg \: ms {}^{ - 1}$

$is \: the \: momentum \: of \: the \: electron.$

(b) Now, using De-broglie formula we have,

p = h/λ

∴ λ = h/p =

$\frac{6.62 \times 10 {}^{ - 34} }{4.02 \times 10 {}^{ - 24} }$

$= 1.64 \times 10 {}^{ - 10}m = 0.164 \times 10 {}^{ - 9m}$

i.e., λ = 0.164 nm , is the De-broglie wavelength of the electron.