Question

What will be the de Broglie wavelength (in metres) of an electron moving with a velocity of 1.2 × 10⁵ m s^–1?

पर्याय

• 6.068 × 10^–9

• 3.133 × 10^–37

• 6.626 × 10^–9

• 6.018 × 10^–35

Answer 1

6.068 × 10^–9 

Explanation:

λ = `h/p`, p = mv

λ = `h/(mv) = (6.626 xx 10^-34)/(9.1 xx 10^-31 xx 1.2 xx 10^5)`

λ = 6.068 × 10^–9 m