A hydrogen atom in its fourth excited state emits a photon with a wavelength of 1282nm. What is the atom’s maximum possible orbital angular momentum (as a multiple of h) after the emission?

Short Answer

Expert verified

Maximum possible orbital angular momentum is 3h2π.

Step by step solution

01

:  Given information

Fourth excited state means n=5

Wavelength, λ=1282nm=1282×10-9m

02

:  Simplification

Using Formula , 1λ=Rz21x2-1n2, where λis the wavelength , Ris rydberg constant , xis the final orbital of electron , nis the initial orbital of electron , zis atomic no

Now for hydrogen z=1

Now , Putting all the values in the formula we get 11282×10-9=1.01×1071x2-152

Solving it we get x=3

It means n=5n=3transition

Now , we know that orbital angular momentum is nh2π

Now after emmision we haven=3

So, the maximum angular momentum comes out to be 3h2π

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