Question

If the electron falls from n=3 to n=2 in hydrogen atom then it's emitted energy is?

Answer 1

SInce the electron makes a transition from n=3 to n=2, we are dealing with a photon emission, which means that the energy difference (ΔE) is actually negative.

ΔE=−RH⋅(1n2f−1n2i), where

nf - the final energy level;
ni - the initial energy level;
RH = 2.18⋅10−18J - Rydberg's constant expressed in J, since energy is expressed in J (this is calculated by multiplying the known R=1.097⋅107m−1 by h⋅c, Planck's constant times the speed of light).

So, the energy difference for a n=3 to n=2 transition is

ΔE=−2.18⋅10−18⋅(122−132)=−2.18⋅10−18⋅0.139=−3.03⋅10−19J