In the Lyman series of hydrogen,which are the two maximum wavelengths? Why?
(N/A) The wavelength $\lambda$ for the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Lyman series,$n_1 = 1$. The wavelength is inversely proportional to the energy difference $\Delta E = E_{n_2} - E_{n_1}$.
To obtain the maximum wavelengths,we need the minimum energy transitions,which correspond to the smallest values of $n_2$ above $n_1$.
For $n_2 = 2$,$\frac{1}{\lambda_1} = R_H (1 - \frac{1}{4}) = \frac{3}{4} R_H$,giving $\lambda_1 \approx 121.6 \ nm$.
For $n_2 = 3$,$\frac{1}{\lambda_2} = R_H (1 - \frac{1}{9}) = \frac{8}{9} R_H$,giving $\lambda_2 \approx 102.6 \ nm$.
Thus,the two maximum wavelengths correspond to transitions from $n_2 = 2$ and $n_2 = 3$ to $n_1 = 1$.
For the Lyman series,$n_1 = 1$. The wavelength is inversely proportional to the energy difference $\Delta E = E_{n_2} - E_{n_1}$.
To obtain the maximum wavelengths,we need the minimum energy transitions,which correspond to the smallest values of $n_2$ above $n_1$.
For $n_2 = 2$,$\frac{1}{\lambda_1} = R_H (1 - \frac{1}{4}) = \frac{3}{4} R_H$,giving $\lambda_1 \approx 121.6 \ nm$.
For $n_2 = 3$,$\frac{1}{\lambda_2} = R_H (1 - \frac{1}{9}) = \frac{8}{9} R_H$,giving $\lambda_2 \approx 102.6 \ nm$.
Thus,the two maximum wavelengths correspond to transitions from $n_2 = 2$ and $n_2 = 3$ to $n_1 = 1$.
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Similar Questions
According to Bohr's theory,
$E_{n} = \text{Total energy}, K_{n} = \text{Kinetic energy}, V_{n} = \text{Potential energy}, r_{n} = \text{Radius of } n^{\text{th}} \text{ orbit}$
Match the following:
Column $I$ Column $II$ $A$. $V_{n} / K_{n} = ?$ $P$. $0$ $B$. If radius of $n^{\text{th}}$ orbit $\propto E_{n}^{x}, x = ?$ $Q$. $-1$ $C$. Angular momentum in lowest orbital $R$. $-2$ $D$. $1/r_{n} \propto Z^{y}, y = ?$ $S$. $1$
$E_{n} = \text{Total energy}, K_{n} = \text{Kinetic energy}, V_{n} = \text{Potential energy}, r_{n} = \text{Radius of } n^{\text{th}} \text{ orbit}$
Match the following:
| Column $I$ | Column $II$ |
| $A$. $V_{n} / K_{n} = ?$ | $P$. $0$ |
| $B$. If radius of $n^{\text{th}}$ orbit $\propto E_{n}^{x}, x = ?$ | $Q$. $-1$ |
| $C$. Angular momentum in lowest orbital | $R$. $-2$ |
| $D$. $1/r_{n} \propto Z^{y}, y = ?$ | $S$. $1$ |
The work function of a metal is $4.2 \, eV$. If radiation of $2000 \, \mathring{A}$ falls on the metal,then the kinetic energy of the fastest photoelectron is:
An electronic transition in a hydrogen atom results in the formation of the $H_\alpha$ line of the hydrogen spectrum in the Lyman series. The energies associated with the electron in each of the orbits involved in the transition (in $kcal \ mol^{-1}$) are:
Threshold frequency,$v_0$ is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency $1.0 \times 10^{15} \ s^{-1}$ was allowed to hit a metal surface,an electron having $1.988 \times 10^{-19} \ J$ of kinetic energy was emitted. Calculate the threshold frequency of this metal. Show that an electron will not be emitted if a photon with a wavelength equal to $600 \ nm$ hits the metal surface.
Medium
View SolutionWhat is the energy associated with the first orbit of $He^{+}$?
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