COMEDK 2024 Shift 2 Paper

To determine the wave number of the longest wavelength transition in the Balmer series of the Hydrogen spectrum, we can start by understanding that the Balmer series refers to electron transitions from higher energy levels to the n=2 level.
The wave number, denoted by ˜ν (nu tilde), is the reciprocal of the wavelength (λ) and can be calculated using the Rydberg formula for the hydrogen spectrum, which is given by:

˜ν=RH(‌

1

n1‌2

−‌

1

n2‌2

)
Here, RH is the Rydberg constant for hydrogen, which is typically:
RH≈109677cm−1
For the Balmer series, n1=2 because all transitions end at the second energy level. The longest wavelength corresponds to the smallest frequency, and thus the smallest energy transition. This occurs when n2 is at the lowest level just above n1, that is, n2=3.
Substituting n1=2 and n2=3 into the Rydberg formula:
˜ν=109677(‌

1

22

−‌

1

32

)
Simplifying the terms inside the parentheses:

‌‌

1

22

=‌

1

4

‌‌

1

32

=‌

1

9

Therefore:
˜ν=109677(‌

1

4

−‌

1

9

)
Finding a common denominator and calculating the difference:
‌

1

4

−‌

1

9

=‌

9−4

36

=‌

5

36

Substituting back into the formula:
˜ν=109677×‌

5

36

Now performing the multiplication:
˜ν=15233cm−1
Thus, the wave number of the longest wavelength transition in the Balmer series of hydrogen is:
Option B: 15233cm−1