Question

if lamda 1 and lambda 2 denote wavelength of de broglie's wave for electron in the first and second bohar orbit in hydrogen atom then ratio of lambda 1 / lambda 2 is​

Answer 1

Answer:

1/4

Explanation:

e=hc/lamda

lamda1 belongs to first orbital energy

Answer 2

Answer:

Explanation:

Given λ₁ is the deBrogile wavelength of electron in the 1st Bohr orbit

and λ₂ the deBrogile wavelength of the electron in the second

deBrogile wavelength = $\frac{h}{p}$ = $\frac{h}{m.v}$

Therefore

λ₁ =$\frac{h}{m.v1}$

λ₂ =$\frac{h}{m.v2}$

Therefore λ₁/λ₂= $\frac{v2}{v1}$

here v₂ is the velocity of the electron in the second orbit

and v₁  is the velocity of the electron in the first orbit

Velocity (v) = 2.18 ×10^(18) × $\frac{n}{Z}$m/s

Therefore

v2/v1 = n2/n1 = 2/1 = 2

Required ratio = 2 : 1