Question

Wavelength of I line of Humphrey series is more than I line of Lyman series in H-atom.
Reason: $\Delta E=\dfrac{hc}{\lambda }$
A. Statement is correct but explanation is incorrect
B. Statement is wrong but explanation is correct
C. Both Statement and explanation are correct and explanation is the correct explanation for Statement.
D. Both Statement and explanation are correct but explanation is not the correct explanation for Statement.

Answer 1

Hint: As we know that Lyman series is seen when the transition of electron takes place from higher energy state$({{n}_{h}}=2,3,4,5...)$ to ${{n}_{1}}$ energy state. Whereas, Humphrey series is seen when the transition of electron takes place from higher energy state $({{n}_{h}}=7,8,9,10,...)$ to ${{n}_{l}}=6$ energy state.

Complete Step by step solution:
- As we know that the relationship between the wavelength and change in energy level is given as:
$\Delta E=\dfrac{hc}{\lambda }$
- It is found that for Humphrey series is:
$\Delta {{E}_{1}}={{E}_{7}}-{{E}_{6}}$
- And for lyman series change in energy can be given as:
$\Delta E={{E}_{2}}-{{E}_{1}}$
- Hence, we can see here that
$\Delta E>\Delta {{E}_{1}}$
- That is the change in the energy level for the lyman series is greater than that for the change in energy level for the Humphrey series.
- And as we know that energy is inversely proportional to the wavelength, therefore wavelength of Humphrey series is greater than that of lyman series (${{\lambda }_{lyman}}<{{\lambda }_{Humphrey}}$)

- Hence, we can conclude that the correct option is (c), that is both Statement and explanation are correct and explanation is the correct explanation for Statement.

Note: - As we know that spectral series are important in astronomical spectroscopy for calculating red shifts and detection of presence of hydrogen.
- We must remember that the Lyman series has lower wavelength compared to all the other spectral series.