Question

A) For hydrogen, find the frequency of light emitted in the transition from the 130th orbit to the 122th orbit.


B) For hydrogen, find the frequency of light absorbed in the transition from the 168th orbit to the 173th orbit.

Answer 1

(A)

here we can use Bohr formula,

1 /  λ = R_H { 1/n₁² - 1/n₂² }   

here , n₁ = 120 , n₂ = 130 , R_H =1.09678 x 10^-2 nm^-1

1 /  λ = R_H { 1/n₁² - 1/n₂² }   

= R_H { 1/ 120² - 1/130²}

= R_H { 1/14400 - 1/16900}

= R_H { 0.0000694444 - 0.00005917159}

= R_H { 0.00001027281}

1 /  λ = 1.09678 x 10^-2 X 0.00001027281

But we know that ,

λ = c/f

c = speed of light

f = frequency

1 /  λ = 1.09678 x 10^-2 X 0.00001027281

1 / (c/f) = 1.09678 x 10^-2 X 0.00001027281

f/c = 1.09678 x 10^-2 X 0.00001027281

f = 1.09678 x 10^-2 X 0.00001027281 x c

f = 1.09678 x 10^-2 X 0.00001027281 x 3 x 10⁸ { c = 3 x 10⁸ }

f = 0.00003380103 x 10⁶

f = 33.80 Hz

(part-B)

1 /  λ = R_H { 1/n₁² - 1/n₂² }   

λ = c/f

hence ,

f/c = R_H { 1/n₁² - 1/n₂² }   

n₁ = 168 , n₂ = 173 , R_H =1.09678 x 10^-2 nm^-1 , C = 3 x 10⁸ m/s

f/c = R_H { 1/168²- 1/173²}

f/c = R_H { 1/28224- 1/29929}

f/c = R_H { 0.00003543083- 0.0000334124}

f/c = R_H { 0.00000201843}

f = R_H x c x { 0.00000201843}

f = 1.09678 x 10^-2 x 3 x 10⁸ x 0.00000201843

f = 0.00000664132 x 10⁶

f = 6.64 Hz