The minimum wavelength of Lyman series lines | vedclass

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Question

(A) For the Lyman series,the wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)

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$\frac{4 P}{3}$

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(A) For the Lyman series,the wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)$,where $n = 2, 3, 4, \dots$$1$. The minimum wavelength $(\lambda_{\min})$ occurs for the transition from $n = \infty$ to $n = 1$:$\frac{1}{\lambda_{\min}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$Given $\lambda_{\min} = P$,we have $P = \frac{1}{R}$.$2$. The maximum wavelength $(\lambda_{\max})$ occurs for the transition from $n = 2$ to $n = 1$:$\frac{1}{\lambda_{\max}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right)$$3$. Substituting $R = \frac{1}{P}$ into the equation:$\frac{1}{\lambda_{\max}} = \frac{1}{P} \cdot \frac{3}{4}$$\lambda_{\max} = \frac{4 P}{3}$.

The minimum wavelength of Lyman series lines is $P$. What is the maximum wavelength of these lines?

Similar Questions

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Let $F_1$ be the frequency of the second line of the Lyman series and $F_2$ be the frequency of the first line of the Balmer series. Then,the frequency of the first line of the Lyman series is given by:

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Assertion: In the Lyman series,the ratio of minimum and maximum wavelength is $\frac{3}{4}$.
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The number of possible spectral lines in a hydrogen atom is:

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