Question

Does a photon of visible light (λ ≈ 400 to 700 nm) have sufficient energy to excite an electron in a hydrogen atom from the n = 1 to the n = 5 energy state? From the n = 2 to the n = 6 energy state?

Answer 1

Answer with Explanation:

We are given that an electron excited from n=1 to n=5 and from n=2 to n=6.

We have to find that the photon of visible light (wavelength 400 to 700 nm) have sufficient energy .

We know that

$\frac{1}{\lambda}=R(\frac{1}{n^2_1}-\frac{1}{n^2_2})$

Where R= Rydberg constant=$1.097\times 10^7/m$

Using the formula

$\frac{1}{\lambda}=1.097\times 10^7(1-\frac{1}{5^2})$

$\frac{1}{\lambda}=1.097\times 10^7(\frac{25-1}{25})=1.097\times 10^7\times \frac{24}{25}$

$\frac{1}{\lambda}=1.05312\times 10^7/m$

$\lambda=\frac{1}{1.05312}\times 10^{-7}=9.5\times 10^{-8}=95\times 10^{-9}=95nm$

$1nm=10^{-9}m$

$\lambda=95nm$

Visible light lies in the region from 400nm-700nm.

It does not lie in this range .Therefore, photon  does not have sufficient energy to excite an electron in hydrogen atom from the n=1 to n=5 energy state.

Now substitute $n_1=2$ and $n_2=6$

$\frac{1}{\lambda}=1.097\times 10^7(\frac{1}{2^2}-\frac{1}{6^2})$

$\frac{1}{\lambda}=1.097\times 10^7(\frac{1}{4}-\frac{1}{36})$

$\frac{1}{\lambda}=1.097\times 10^7(\frac{9-1}{36})$

$\frac{1}{\lambda}=1.097\times 10^7\times \frac{8}{36}=0.244\times 10^7$

$\lambda=\frac{1}{0.244}\times 10^{-7}=4.10\times 10^{-7}m$

$\lambda=410\times 10^{-9}m=410 nm$

It lies in the given range.

Therefore, a photon of visible light have sufficient energy to excite an electron in a hydrogen atom from energy state n=2 to n=6.

Answer 2

The excitation from n = 2 to the n = 6 is achieved by a photon of wavelength 410 nm in the visible spectrum.

Using the Rydberg relation;

1/λ = R(1/nf^2 - 1/ni^2)

Where;

R = 1.097 × 10^7 m-1

nf = 5

ni = 1

1/λ = 1.097 × 10^7(1/1^2 - 1/5^2)

λ = 94.9 nm

A photon of visible light can not perform this excitation.

R = 1.097 × 10^7 m-1

nf = 6

ni = 2

1/λ = 1.097 × 10^7(1/2^2 - 1/6^2)

λ = 410nm

This lies within the visible spectrum and is achieved by visible light.

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