Question

The wavelength of electrons accelerated from rest through a potential difference of 40 kV is X x 10^–12 m. The value of x is ....... (Nearest integer) 

Given : Mass of electron = 9.1 x 10^–31 kg 

Charge on an electron = 1.6 x 10^–19 C 

Planck's constant = 6.63 x 10^–34 Js

Answer 1

Wavelength (λ) of electron is given by

\(\lambda=\frac{h}{\sqrt{2mqV}}\)

Here

h is planck's constant

m is mass of electron

q is harge on electron

V is potential difference

Putting the values:

\(\lambda=\frac{6.63\times 10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times 1.6\times 10^{-19}\times 40\times 10^3}}\)

\(=\frac{6.63\times10^{-34}}{\sqrt{116.48\times10^{-46}}}\) 

\(= 6.144\times10^{-12}=6\times10^{-12}\)

x = 6​​​​​​​​​​​​​​

Answer 2

Answer is (6)

De-broglie-wave length of electron:

= 0.614 x 10^–11 m 

= 6.14 × 10^–12 m 

Nearest integer = 6 

OR

λ = \(\frac{12.3}{\sqrt{V}}\)Å

= \(\frac{12.3}{200}\) = 6.15 x 10^-12m

Ans is 6