Question

The frequency of radiation emitted when an electron falls from n=4 to n=1 in a H atom will be ( Given that: ionisation energy of H atom : 2.18 x 10^ -18 J/atom and Planck's constant = 6.625 x 10^ -34 J/s). :
a) 1.03 x 10^15 per second. b) 3.08 x 10^ 15 per second
c) 2.00 x 10^ 15 per second. d) 1.54 x 10^15 per second

Answer 1

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Answer 2

Frequency and energy both are related to each other and this relation can be write is as-

Δ E = hv where E is the energy, v which is actually nu is frequency and h is the Plank’s constant.

We can also write is as, v = ΔE/h

$\Delta E=2.18 \times 10^{-18} \times\left[1 / n 1^{2}-1 / n 2^{2}\right]$

$\Delta E=2.18 \times 10-18 \times\left[1 / 1^{2}-1 / 4^{2}\right]$ (value of n is given in the question)

$\Delta E=2.18 \times 10^{-18} \times 15 / 16$

Now put this value of ΔE in the above mention equation v= ΔE/h

$\mathrm{v}=2.18 \times 10^{-18} \times 15 / 16 / 6.625 \times 10^{-34}=0.308 \times 10^{16}$

$\bold{\mathrm{v}=3.08 \times 10^{15} / \mathrm{s}}$