1.

What are the frequency and wavelength of a photon emitted during a transition from n=5 state to n=2 state in hydrogen atom???

Answer»

n^1 = lower energy STATE, i.e 2 in this question.

n^2 = HIGHER energy state, i.e 5

Z = atomic number, here Z for hydrogen is 1

On substituting the given values we get,

1/∆ = 1.097 x 10^7(1/2^2–1/5^2)1^2

1/∆ = 1.097x 10^7(1/4–1/25)

1/∆ = 1.097 x 10^7(0.21)

1/∆ = 0.23 x 10^7

On doing reciprocal we get ∆(lambda), i.e the wavelength of the emitted photon.

∆= 4.347 x 10^-7 m

But we require its FREQUENCY,

Therefore, frequency = C/∆

(c= speed of light)

3x10^8/4.347x10^-7 = 0.69 x 10^15

Therefore, frequency= 0.69 x 10^15

Hope it helps!
Thank You!


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