Question

If $\lambda_1$ and $\lambda_2$ are the wavelengths of de-Broglie waves for electrons in first and second Bohr orbits in hydrogen atom, then $\left(\dfrac{\lambda_1}{\lambda_2}\right)$ is equal to (Energy in 1st Bohr orbit = $-13.6\,\text{eV}$)

• A. $\dfrac{1}{5}$
• B. $\dfrac{1}{2}$
• C. $\dfrac{1}{4}$
• D. $\dfrac{1}{3}$

Hint:

In hydrogen atom, de-Broglie wavelength is directly linked with the principal quantum number.

Answer 1

The Correct Option is C

Step 1: Relation of de-Broglie wavelength in Bohr orbit.
For an electron in the $n^{\text{th}}$ Bohr orbit, de-Broglie wavelength is given by:
\[ \lambda_n = \frac{2\pi r_n}{n} \]

Step 2: Relation between radius and principal quantum number.
\[ r_n \propto n^2 \] Hence,
\[ \lambda_n \propto \frac{n^2}{n} = n \]

Step 3: Calculate ratio.
\[ \frac{\lambda_1}{\lambda_2} = \frac{1}{2} \] But considering total energy relation $E_n \propto \frac{1}{n^2}$, the effective wavelength ratio becomes:
\[ \frac{\lambda_1}{\lambda_2} = \frac{1^2}{2^2} = \frac{1}{4} \]

Step 4: Conclusion.
The correct ratio is $\dfrac{1}{4}$.