Question

Which of the following transition in the hydrogen atom
will fall in visible spectrum?
n = 4 to n = 3
n = 2 to n = 1
n = 3 to n = 2
n = 5 to n = 4​

Answer 1

It's N = 5 to N = 4

Large enough to be visible in spectrum.... Lower than this u can't see it there...

Answer 2

Option (c) , n=3 to n=2 is the correct answer.

• The possible lines when a hydrogen atom undergoes transition are LYMAN, BALMER, PASCHEN, BRACKET, PFUND, HUMPHREY.

• Lyman series occurs in the ultraviolet region. When an electron jumps from a higher energy level with a value of n>=2 to a lower energy level having a value of n=1, the group of lines produced is called the Lyman series.

• Balmer series occurs in the Visible Region. When an electron jumps from a higher energy level with a value of n>=3 to a lower energy level having a value of n=2, the group of lines produced is called the Balmer series.

• Paschen series occurs in the Infrared Region. When an electron jumps from a higher energy level with a value of n>=4 to a lower energy level having a value of n=3, the group of lines produced is called the Paschen series.

• Brackett series occurs in the Infrared region. When an electron jumps from a higher energy level with a value of n>=5 to a lower energy level having a value of n=4, the group of lines produced is called the Brackett series.

• Pfund series occurs in the Infrared region. When an electron jumps from a higher energy level with a value of n>=6 to a lower energy level having a value of n=5, the group of lines produced is called the Paschen series.

• Humphrey series occurs in the Far Infrared region. When an electron jumps from a higher energy level with a value of n>=7 to a lower energy level having a value of n=6, the group of lines produced is called the Humphrey series.

For a hydrogen atom to fall in the visible region, the transition should be of the Balmer series.

• n = 4 to n = 3 is a Paschen series transition.

• n = 2 to n = 1 is  a Lyman series transition.

• n = 3 to n = 2 is Balmer series transition.

• n = 5 to n = 4 is Brackett series transition.

Hence, The line n=3 to n=2 will fall in the visible spectrum as it belongs to the Balmer series.

Therefore, Option (c) is the correct answer.