Question

What is the wavelength of photon emitted during electron transition from n=5 to n=3 in the state of hydrogen atom
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Answer 1

Explanation:

According to your question it is given that a electron in n=5 goes to n=3 in a hydrogen atom and some wavelength is emitted. We have to find that wavelength. Right?

In that case what I should do? I only know the shell number.

The procedure is that we have to find the amount of energy involved during this activity. I still don't know if the energy is released or absorbed , but do not need to worry , the math will take care of that. using relation : ∆E = E5 - E3 where E5 and E3 are energies of 5th and 3rd shells respectively.

Now once I find the energy I can use Planck's equation: ∆E = h√

where ∆E = energy involved.

h = Planck's constant

√ = frequency of radiation

From here I can use relation lamda = c/ √

where lamda = wavelength of radiation

c = speed of light

√ = frequency of radiation.

quite a lengthy procedure .

So here we have a shortcut ,

thanks to Rydberg Equation we can directly calculate the wavelength

(1/lamda) = RH × Z^2 × { 1/ (n1^2) - 1/ ( n2^2) }

plugging the values of n1 = 3 and n2 = 5

RH = Rydberg constant ~ 10^(-7)

Z = 1 for hydrogen atom

we get this:

(1/ lamda) = 10^(-7) × 20/ 225

lamda = 225/( 20× 10^-7)

{lamda = 11.25 × 10^7}

Therefore wavelength emitted should be (11.25 × 10^7) m

* note: if my answer is wrong please let me know , also the procedure is same.

Hope it helps.