If $\lambda_1$ is the wavelength of the series limit of the Lyman series,$\lambda_2$ is the wavelength of the first line of the Lyman series,and $\lambda_3$ is the series limit of the Balmer series,then the relation between $\lambda_1, \lambda_2,$ and $\lambda_3$ is:

  • A
    $\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \frac{1}{\lambda_3}$
  • B
    $\frac{1}{\lambda_1} = \frac{1}{\lambda_2} - \frac{1}{\lambda_3}$
  • C
    $\lambda_2 = \lambda_1 + \lambda_3$
  • D
    $\lambda_1 = \lambda_2 + \lambda_3$

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If $R$ is the Rydberg constant for hydrogen,the wave number of the first line in the Lyman series will be

The energy levels of the hydrogen spectrum are shown in the figure. There are some transitions $A, B, C, D$ and $E$. Transitions $A, B$ and $C$ respectively represent:

Match List $I$ with List $II$.
List $I$ (Spectral Lines of Hydrogen for transitions from) List $II$ (Wavelengths $(nm)$)
$A$. $n_2=3$ to $n_1=2$ $I$. $410.2$
$B$. $n_2=4$ to $n_1=2$ $II$. $434.1$
$C$. $n_2=5$ to $n_1=2$ $III$. $656.3$
$D$. $n_2=6$ to $n_1=2$ $IV$. $486.1$

Choose the correct answer from the options given below:

The shortest wavelength for the Lyman series is $912 \ \text{Å}$. The longest wavelength in the Paschen series is: (in $\text{Å}$)

The ratio of the wavelengths of the first Lyman line and the second Balmer line of the hydrogen atom is

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