Question

In which of the following Bohr’s orbit (n) a hydrogen atom emits the photons of lowest frequency?

• A. n = 2 to n = 1
• B. n = 4 to n = 2
• C. n = 4 to n = 1
• D. n = 4 to n = 3

Answer 1

Answer: (D)

n = 4 to n = 3

Answer 2

The energy level diagram schematically looks like this as we move to higher and higher levels, the energy increase but the energy gap between any two levels decreases hence in the transition from n = 4 to n = 3energy released will be $E_{4} - E_{3}$

And in the transition from n = 2 to n = 1 energy released is $E_{2} - E_{1}$

Now as $E_{4} - E_{3} < E_{2} - E_{1}$

$\Rightarrow h\upsilon_{43} < h\upsilon_{21} \Rightarrow \upsilon_{43} < \upsilon_{21}$