Question

A ground state hydrogen atom absorbs a photon of light having a wavelength of 93.03 nm. it then gives off a photon having a wavelength of 2165 nm. What is the final state of the hydrogen atom?

Answer 1

We know that,

E=hc/wavelength

From the above formula we will first find the value of E.

So,E=(6.626×10^-34 j.s ×3×10^8 m/s)/(93.03×10^-9 m)

=2.13673 × 10^-18 j.

Now putting the value of E in rydberg equation as follow,:

E=2.18× 10^--18(1/1 - 1/n^2 )

Substituting the value of E to find n,

2.13673× 10^-18 j =2.18 ×10^-18 j (1/1 - 1/n^2 )

n^2= 1/.020183

n=7

Now we have find the value of E for 2165nm.

E=hc/wavelength

E=(6.626×10^-34 j.s ×3×10^8 m/s)/(2165×10^-9 m)

E=9.18152×10^-20 j

Now substituting the value of E in rydberg equation to find n,

9.18152×10^-20 j =2.18×10^-18 j(1/n^2 -1/7^2)

n =4

So when the value of n is 4 it released a photon of 2165nm.

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