Q23 / 75 · JEE Main 2026 (08 April Shift 2) · Chemistry
Consider two radiations of wavelengths
1. $\lambda_1 = 2000$ Å
2. $\lambda_2 = 6000$ Å
The ratio of the energies of these two radiations $\left(\dfrac{E_1}{E_2}\right)$ is ________ (Nearest integer).
1. $\lambda_1 = 2000$ Å
2. $\lambda_2 = 6000$ Å
The ratio of the energies of these two radiations $\left(\dfrac{E_1}{E_2}\right)$ is ________ (Nearest integer).
Correct answer: 3
Solution
The energy of a photon is given by the relation:
$E = \dfrac{hc}{\lambda}$
For the two radiations, the energies are $E_1 = \dfrac{hc}{\lambda_1}$ and $E_2 = \dfrac{hc}{\lambda_2}$.
The ratio of their energies is:
$\dfrac{E_1}{E_2} = \dfrac{\lambda_2}{\lambda_1}$
Substituting the given values $\lambda_1 = 2000$ and $\lambda_2 = 6000$:
$\dfrac{E_1}{E_2} = \dfrac{6000}{2000} = 3$
Answer: $3$
$E = \dfrac{hc}{\lambda}$
For the two radiations, the energies are $E_1 = \dfrac{hc}{\lambda_1}$ and $E_2 = \dfrac{hc}{\lambda_2}$.
The ratio of their energies is:
$\dfrac{E_1}{E_2} = \dfrac{\lambda_2}{\lambda_1}$
Substituting the given values $\lambda_1 = 2000$ and $\lambda_2 = 6000$:
$\dfrac{E_1}{E_2} = \dfrac{6000}{2000} = 3$
Answer: $3$
Hence, the correct answer is 3.