Question

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

Answer 1

here is the answer...hope it helps

Answer 2

The frequency and wavelength of a photon emitted during the transition from the n=5 state to the n=2 state in the hydrogen atom are 6.9× 10^14 s-1 and 435nm.

Given:

A photon is emitted during the transition from the n=5 state to the n=2 state in the hydrogen atom.

To Find:

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

Solution:

To find the frequency and wavelength of a photon emitted we will follow the following steps:

Energy is calculated as

$\frac{13.6}{ {n}^{2} }$

The energy at n= 5,

$\frac{13.6}{ {5}^{2} } = \frac{13.6}{25} = 0.544ev = 0.544×1.602×10^-19= 0.87×10^-19$

The energy at n = 2

$\frac{13.6}{ {2}^{2} } = \frac{13.6}{4} =3.4ev = 3.4×1.602×10^-19= 5.44×10^-19$

The energy difference between the two levels is E2 - E5 = (5.44-0.87) × 10^-19 = 4.57× 10^-19J

Now,

Frequency =

$\frac{E}{h}$

h is plank's constant = 6.63× 10^-34

Frequency =

$\frac{4.57× 10^-19}{6.63× 10^-34}= 6.9×10^14 s-1$

Wavelength =

$\frac{speed}{frequency} = \frac{3×10^8}{6.9×10^14 } = 435×10^-9m = 435nm$

Henceforth, the frequency and wavelength of a photon emitted during the transition from the n=5 state to the n=2 state in the hydrogen atom are 6.9× 10^14 s-1 and 435nm.

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