Q148 / 150 · MHT CET 2026 (17 April Shift 1) · Physics
For wavelength of visible radiation of Hydrogen spectrum, Balmer gave an equation as $\lambda = \dfrac{xm^2}{m^2-4}$, where m is the integer value. The value of x in terms of Rydberg's constant R is
- A$\dfrac{R}{4}$
- Correct answerB$\dfrac{4}{R}$
- C$2R$
- D$4R$
Solution
The Rydberg formula for the Balmer series of the hydrogen spectrum is given by
$\dfrac{1}{\lambda} = R \left( \dfrac{1}{2^2} - \dfrac{1}{m^2} \right)$
where $m$ is an integer greater than $2$.
Simplifying the expression, we get
$\dfrac{1}{\lambda} = R \left( \dfrac{m^2 - 4}{4m^2} \right)$
$\lambda = \dfrac{4m^2}{R(m^2 - 4)}$
$\lambda = \dfrac{\left(\dfrac{4}{R}\right)m^2}{m^2 - 4}$
Comparing this with the given equation $\lambda = \dfrac{xm^2}{m^2 - 4}$, we get
$x = \dfrac{4}{R}$
Answer: $\dfrac{4}{R}$
$\dfrac{1}{\lambda} = R \left( \dfrac{1}{2^2} - \dfrac{1}{m^2} \right)$
where $m$ is an integer greater than $2$.
Simplifying the expression, we get
$\dfrac{1}{\lambda} = R \left( \dfrac{m^2 - 4}{4m^2} \right)$
$\lambda = \dfrac{4m^2}{R(m^2 - 4)}$
$\lambda = \dfrac{\left(\dfrac{4}{R}\right)m^2}{m^2 - 4}$
Comparing this with the given equation $\lambda = \dfrac{xm^2}{m^2 - 4}$, we get
$x = \dfrac{4}{R}$
Answer: $\dfrac{4}{R}$
Hence, the correct answer is option: 'B'.