For hydrogen atom, 1 and 2 are the wavelengths corresponding to t… — Solution | JEE Physics

NVAEasyJEE 2023Bohr's Model & Hydrogen Spectrum

For hydrogen atom, $\lambda_1$ and $\lambda_2$ are the wavelengths corresponding to the transitions $1$ and $2$ , respectively, as shown in the figure. The ratio of $\lambda_1$ and $\lambda_2$ is $\frac{x}{32}$ . The value of $x$ is:

Energy level diagram of hydrogen atom with levels $$n=1$$, $$n=2$$, and $$n=3$$, showing transition $$1$$ from $$n=3$$ to $$n=1$$ and transition $$2$$ from $$n=2$$ to $$n=1$$.

Answer

Correct answer:27

Step-by-step solution

Standard Method

Given: Two hydrogen atom transitions produce wavelengths $\lambda_1$ and $\lambda_2$ as shown in the figure.

Find: The value of $x$ in $\frac{\lambda_1}{\lambda_2} = \frac{x}{32}$ .

Use the Rydberg formula for hydrogen:

\frac{1}{\lambda} = R_z \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

For transition $1$ :

\frac{1}{\lambda_1} = R_z \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R_z \left( 1 - \frac{1}{9} \right) = \frac{8}{9} R_z

For transition $2$ :

\frac{1}{\lambda_2} = R_z \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_z \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R_z

Now,

\frac{\lambda_1}{\lambda_2} = \frac{\frac{1}{\left( \frac{8}{9} R_z \right)}}{\frac{1}{\left( \frac{3}{4} R_z \right)}} = \frac{9}{8} \times \frac{3}{4} = \frac{27}{32}

Comparing with $\frac{x}{32}$ , we get $x = 27$ .

Therefore, the value of $x$ is $27$ .

Using the extracted working

Given: the solution states the correct answer is $27$ .

Find: Verify the ratio using the transition values shown.

From the extracted working:

\frac{1}{\lambda_1} = \frac{9}{8} R_z

and

\frac{1}{\lambda_2} = \frac{4}{3} R_z

Hence,

\frac{\lambda_1}{\lambda_2} = \frac{27}{32}

So, in the form $\frac{x}{32}$ , we obtain $x = 27$ .

Therefore, the numerical answer is $27$ .

Common mistakes

Using the ratio of inverse wavelengths directly as the ratio of wavelengths. This is wrong because $\lambda$ is inversely proportional to the Rydberg expression. After finding $\frac{1}{\lambda}$ , invert before comparing $\lambda_1$ and $\lambda_2$ .
Reading the energy-level diagram incorrectly. This is wrong because the initial and final levels determine the value of $n_1$ and $n_2$ in the Rydberg formula. First identify each transition carefully from the figure, then substitute the quantum numbers.
Substituting the higher level first and lower level second without maintaining the emitted-photon convention. This can give a negative value for $\frac{1}{\lambda}$ . Use the lower final level term minus the higher initial level term so that the wavelength remains positive.

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Step-by-step solution

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