Question

A proton is accelerated through a potential difference of 400V to have same de broglie wavelength what potential diffetence must be applied across doubly ionised 8O16 atom

Answer 1

The potential difference is V1 = 8.34 x 10^-31 volts

Explanation:

De-Broglie wavelength (λ) is given by

λ=h / mv

⇒ λ = h / √2mqV

For proton:

λ =  h /  √2 x 1.67 x 10^-31 x 1.6 x 10^-19 x 400

For oxygen atom:

λ1 =  h /  √2 x 8 x 1.6 x 10^-19 x 16 x v1

For same wavelength:

λ  = λ 1

V1 = 2 x 1.67 x x 10^-31  x 1.6 x 10^-19 x 400

V1 = 8.34 x 10^-31 volts

Thus the potential difference is V1 = 8.34 x 10^-31 volts