Question

For what k.e of neutron will the associated de-Broglie wavelength be 1.40 × 10^-10 m? mass of neutron= 1.675 × 10^-27 kg, h = 6.63 × 10^-34 J

पर्याय

• 6.634 × 10^-21 J

• 4.262 × 10^-19 J

• 1.2 × 10^-10 m

• 2.5 × 10^-19 J

Answer 1

6.634 × 10^-21 J

Explanation:

`lambda = h/sqrt(2ME)` or E = `h^2/(2lambda2m)`

= `(6.63 xx 10^-34)^2/(2 xx (1.4 xx 10^-10)^2 xx (1.674 xx 10)^-27`

= 6.634 × 10^-21 J.