Given 3p orbital In $3 \mathrm{p}$ orbital, $\mathrm{n}=3$ and $l=1$ We know, $\text { Spherical Node }=\mathrm{n}-l-1=3-1-1=1$ Non spherical Node $=l=1 Hence, one spherical Node and one non spherical Node
J and K CET-(1998)
Structure of Atom
238840
Among the first lines of Lyman, Balmer, Paschen and Brackett series in hydrogen atomic spectra, which has the highest energy?
1 Lyman
2 Balmer
3 Paschen
4 Brackett
Explanation:
Energy, $\Delta \mathrm{E}=13.6\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \mathrm{eV}$ For the first line of Lyman series, $\begin{aligned} & \mathrm{n}_1=1, \mathrm{n}_2=2 \\ & \Delta \mathrm{E}=13.6\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \mathrm{eV}=10.2 \mathrm{eV} \end{aligned}$ and energy decreases as we move on to the next series.
AP-EAMCET-1999
Structure of Atom
238842
The wavelength of spectral line emitted by hydrogen atom in the Lyman series is $\frac{16}{15 R} \mathrm{~cm}$. What is the value of $n_2$ ? (R=Rydberg constant)
1 2
2 3
3 4
4 1
Explanation:
Given, $\lambda=\frac{16}{15 \mathrm{R}} \mathrm{cm}, \mathrm{n}_2=$ ? For Lyman series- $\begin{array}{ll} & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ & \frac{15 \mathrm{R}}{16}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ \text { or } \quad \frac{15 \mathrm{R}}{16 \mathrm{R}}=1-\frac{1}{\mathrm{n}_2^2} \\ \text { or } & \frac{1}{\mathrm{n}_2^2}=1-\frac{15}{16}=\frac{1}{16} \\ \text { or } & \frac{1}{\mathrm{n}_2}=\frac{1}{4} \\ \text { or } & \mathrm{n}_2=4 \end{array}$
AP-EAMCET-2007
Structure of Atom
238843
An electronic transition in hydrogen atom results in the formation of $\mathbf{H}_\alpha$ line of hydrogen in Lyman series, the energies associated with the electron in each of the orbits involved in the transition (in kcal $\mathbf{~ m o l}^{-1}$ ) are :
1 $-313.6,-34.84$
2 $-313.6,-78.4$
3 $-78.4,-34.84$
4 $-78.4,-19.6$
Explanation:
Energy of electron in $\mathrm{n}^{\text {th }}$ orbit- $\mathrm{E}_{\mathrm{n}}=\frac{2 \pi^2 Z^2 \mathrm{e}^4}{\mathrm{n}^2 \mathrm{~h}^2} \mathrm{a}$ Putting the value of $\mathrm{h}$, e and $\pi$, we get $\begin{aligned} \mathrm{E}_{\mathrm{n}} & =\frac{-1311.8 \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{~kJ} / \mathrm{mole} \\ \mathrm{E}_{\mathrm{n}} & =\frac{-313.52 \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{kcal} / \mathrm{mole} \\ & (\because 1 \mathrm{kcal}=4.184 \mathrm{~kJ}) \end{aligned}$ For $\mathrm{H}$ atom, $\mathrm{Z}=1$ and Lyman series have the value of $\mathrm{n}_1=1, \mathrm{n}_2=2$. $\mathrm{n}_1=1, \mathrm{n}_2=2$ Since, Energy of electron in $\mathrm{n}_1$ orbit $\begin{aligned} & =\frac{-313.52 \times(1)^2}{(1)^2} \\ & =-313.6 \mathrm{kcal} / \mathrm{mole} \end{aligned}$ and energy of electron in $\mathrm{n}_2$ orbit $\begin{aligned} & =\frac{-313.52 \times(1)^2}{(2)^2} \\ & =-78.38 \mathrm{kcal} / \mathrm{mole} \end{aligned}$
AP-EAMCET-2008
Structure of Atom
238844
The ratio of the highest to the lowest wavelength of Lyman series is
1 $4: 3$
2 $9: 8$
3 $27: 5$
4 $16: 5$
Explanation:
The equation of Lyman series wavelength for hydrogen atom is $\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}}\left(1-\frac{1}{\mathrm{n}^2}\right)$ Where, $\lambda=$ Lyman series wavelength for hydrogen atom $mathrm{R}_{\mathrm{H}}=\text { Rydberg Constant }$ For lowest wavelength, $\mathrm{n}=\infty$ $\therefore \frac{1}{\lambda_L}=\mathrm{R}_{\mathrm{H}}\left(1-\frac{1}{\infty^2}\right)$ or $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}_{\mathrm{H}}$ For highest wavelength, $\mathrm{n}=2$ $\therefore \frac{1}{\lambda_{\mathrm{H}}}=\mathrm{R}_{\mathrm{H}}\left(1-\frac{1}{4}\right)=\frac{3 \mathrm{R}_{\mathrm{H}}}{4}$ From equation (1) and (2) $\frac{\lambda_{\mathrm{H}}}{\lambda_{\mathrm{L}}}=\frac{4}{3} \Rightarrow 4: 3$
Given 3p orbital In $3 \mathrm{p}$ orbital, $\mathrm{n}=3$ and $l=1$ We know, $\text { Spherical Node }=\mathrm{n}-l-1=3-1-1=1$ Non spherical Node $=l=1 Hence, one spherical Node and one non spherical Node
J and K CET-(1998)
Structure of Atom
238840
Among the first lines of Lyman, Balmer, Paschen and Brackett series in hydrogen atomic spectra, which has the highest energy?
1 Lyman
2 Balmer
3 Paschen
4 Brackett
Explanation:
Energy, $\Delta \mathrm{E}=13.6\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \mathrm{eV}$ For the first line of Lyman series, $\begin{aligned} & \mathrm{n}_1=1, \mathrm{n}_2=2 \\ & \Delta \mathrm{E}=13.6\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \mathrm{eV}=10.2 \mathrm{eV} \end{aligned}$ and energy decreases as we move on to the next series.
AP-EAMCET-1999
Structure of Atom
238842
The wavelength of spectral line emitted by hydrogen atom in the Lyman series is $\frac{16}{15 R} \mathrm{~cm}$. What is the value of $n_2$ ? (R=Rydberg constant)
1 2
2 3
3 4
4 1
Explanation:
Given, $\lambda=\frac{16}{15 \mathrm{R}} \mathrm{cm}, \mathrm{n}_2=$ ? For Lyman series- $\begin{array}{ll} & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ & \frac{15 \mathrm{R}}{16}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ \text { or } \quad \frac{15 \mathrm{R}}{16 \mathrm{R}}=1-\frac{1}{\mathrm{n}_2^2} \\ \text { or } & \frac{1}{\mathrm{n}_2^2}=1-\frac{15}{16}=\frac{1}{16} \\ \text { or } & \frac{1}{\mathrm{n}_2}=\frac{1}{4} \\ \text { or } & \mathrm{n}_2=4 \end{array}$
AP-EAMCET-2007
Structure of Atom
238843
An electronic transition in hydrogen atom results in the formation of $\mathbf{H}_\alpha$ line of hydrogen in Lyman series, the energies associated with the electron in each of the orbits involved in the transition (in kcal $\mathbf{~ m o l}^{-1}$ ) are :
1 $-313.6,-34.84$
2 $-313.6,-78.4$
3 $-78.4,-34.84$
4 $-78.4,-19.6$
Explanation:
Energy of electron in $\mathrm{n}^{\text {th }}$ orbit- $\mathrm{E}_{\mathrm{n}}=\frac{2 \pi^2 Z^2 \mathrm{e}^4}{\mathrm{n}^2 \mathrm{~h}^2} \mathrm{a}$ Putting the value of $\mathrm{h}$, e and $\pi$, we get $\begin{aligned} \mathrm{E}_{\mathrm{n}} & =\frac{-1311.8 \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{~kJ} / \mathrm{mole} \\ \mathrm{E}_{\mathrm{n}} & =\frac{-313.52 \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{kcal} / \mathrm{mole} \\ & (\because 1 \mathrm{kcal}=4.184 \mathrm{~kJ}) \end{aligned}$ For $\mathrm{H}$ atom, $\mathrm{Z}=1$ and Lyman series have the value of $\mathrm{n}_1=1, \mathrm{n}_2=2$. $\mathrm{n}_1=1, \mathrm{n}_2=2$ Since, Energy of electron in $\mathrm{n}_1$ orbit $\begin{aligned} & =\frac{-313.52 \times(1)^2}{(1)^2} \\ & =-313.6 \mathrm{kcal} / \mathrm{mole} \end{aligned}$ and energy of electron in $\mathrm{n}_2$ orbit $\begin{aligned} & =\frac{-313.52 \times(1)^2}{(2)^2} \\ & =-78.38 \mathrm{kcal} / \mathrm{mole} \end{aligned}$
AP-EAMCET-2008
Structure of Atom
238844
The ratio of the highest to the lowest wavelength of Lyman series is
1 $4: 3$
2 $9: 8$
3 $27: 5$
4 $16: 5$
Explanation:
The equation of Lyman series wavelength for hydrogen atom is $\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}}\left(1-\frac{1}{\mathrm{n}^2}\right)$ Where, $\lambda=$ Lyman series wavelength for hydrogen atom $mathrm{R}_{\mathrm{H}}=\text { Rydberg Constant }$ For lowest wavelength, $\mathrm{n}=\infty$ $\therefore \frac{1}{\lambda_L}=\mathrm{R}_{\mathrm{H}}\left(1-\frac{1}{\infty^2}\right)$ or $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}_{\mathrm{H}}$ For highest wavelength, $\mathrm{n}=2$ $\therefore \frac{1}{\lambda_{\mathrm{H}}}=\mathrm{R}_{\mathrm{H}}\left(1-\frac{1}{4}\right)=\frac{3 \mathrm{R}_{\mathrm{H}}}{4}$ From equation (1) and (2) $\frac{\lambda_{\mathrm{H}}}{\lambda_{\mathrm{L}}}=\frac{4}{3} \Rightarrow 4: 3$
Given 3p orbital In $3 \mathrm{p}$ orbital, $\mathrm{n}=3$ and $l=1$ We know, $\text { Spherical Node }=\mathrm{n}-l-1=3-1-1=1$ Non spherical Node $=l=1 Hence, one spherical Node and one non spherical Node
J and K CET-(1998)
Structure of Atom
238840
Among the first lines of Lyman, Balmer, Paschen and Brackett series in hydrogen atomic spectra, which has the highest energy?
1 Lyman
2 Balmer
3 Paschen
4 Brackett
Explanation:
Energy, $\Delta \mathrm{E}=13.6\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \mathrm{eV}$ For the first line of Lyman series, $\begin{aligned} & \mathrm{n}_1=1, \mathrm{n}_2=2 \\ & \Delta \mathrm{E}=13.6\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \mathrm{eV}=10.2 \mathrm{eV} \end{aligned}$ and energy decreases as we move on to the next series.
AP-EAMCET-1999
Structure of Atom
238842
The wavelength of spectral line emitted by hydrogen atom in the Lyman series is $\frac{16}{15 R} \mathrm{~cm}$. What is the value of $n_2$ ? (R=Rydberg constant)
1 2
2 3
3 4
4 1
Explanation:
Given, $\lambda=\frac{16}{15 \mathrm{R}} \mathrm{cm}, \mathrm{n}_2=$ ? For Lyman series- $\begin{array}{ll} & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ & \frac{15 \mathrm{R}}{16}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ \text { or } \quad \frac{15 \mathrm{R}}{16 \mathrm{R}}=1-\frac{1}{\mathrm{n}_2^2} \\ \text { or } & \frac{1}{\mathrm{n}_2^2}=1-\frac{15}{16}=\frac{1}{16} \\ \text { or } & \frac{1}{\mathrm{n}_2}=\frac{1}{4} \\ \text { or } & \mathrm{n}_2=4 \end{array}$
AP-EAMCET-2007
Structure of Atom
238843
An electronic transition in hydrogen atom results in the formation of $\mathbf{H}_\alpha$ line of hydrogen in Lyman series, the energies associated with the electron in each of the orbits involved in the transition (in kcal $\mathbf{~ m o l}^{-1}$ ) are :
1 $-313.6,-34.84$
2 $-313.6,-78.4$
3 $-78.4,-34.84$
4 $-78.4,-19.6$
Explanation:
Energy of electron in $\mathrm{n}^{\text {th }}$ orbit- $\mathrm{E}_{\mathrm{n}}=\frac{2 \pi^2 Z^2 \mathrm{e}^4}{\mathrm{n}^2 \mathrm{~h}^2} \mathrm{a}$ Putting the value of $\mathrm{h}$, e and $\pi$, we get $\begin{aligned} \mathrm{E}_{\mathrm{n}} & =\frac{-1311.8 \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{~kJ} / \mathrm{mole} \\ \mathrm{E}_{\mathrm{n}} & =\frac{-313.52 \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{kcal} / \mathrm{mole} \\ & (\because 1 \mathrm{kcal}=4.184 \mathrm{~kJ}) \end{aligned}$ For $\mathrm{H}$ atom, $\mathrm{Z}=1$ and Lyman series have the value of $\mathrm{n}_1=1, \mathrm{n}_2=2$. $\mathrm{n}_1=1, \mathrm{n}_2=2$ Since, Energy of electron in $\mathrm{n}_1$ orbit $\begin{aligned} & =\frac{-313.52 \times(1)^2}{(1)^2} \\ & =-313.6 \mathrm{kcal} / \mathrm{mole} \end{aligned}$ and energy of electron in $\mathrm{n}_2$ orbit $\begin{aligned} & =\frac{-313.52 \times(1)^2}{(2)^2} \\ & =-78.38 \mathrm{kcal} / \mathrm{mole} \end{aligned}$
AP-EAMCET-2008
Structure of Atom
238844
The ratio of the highest to the lowest wavelength of Lyman series is
1 $4: 3$
2 $9: 8$
3 $27: 5$
4 $16: 5$
Explanation:
The equation of Lyman series wavelength for hydrogen atom is $\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}}\left(1-\frac{1}{\mathrm{n}^2}\right)$ Where, $\lambda=$ Lyman series wavelength for hydrogen atom $mathrm{R}_{\mathrm{H}}=\text { Rydberg Constant }$ For lowest wavelength, $\mathrm{n}=\infty$ $\therefore \frac{1}{\lambda_L}=\mathrm{R}_{\mathrm{H}}\left(1-\frac{1}{\infty^2}\right)$ or $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}_{\mathrm{H}}$ For highest wavelength, $\mathrm{n}=2$ $\therefore \frac{1}{\lambda_{\mathrm{H}}}=\mathrm{R}_{\mathrm{H}}\left(1-\frac{1}{4}\right)=\frac{3 \mathrm{R}_{\mathrm{H}}}{4}$ From equation (1) and (2) $\frac{\lambda_{\mathrm{H}}}{\lambda_{\mathrm{L}}}=\frac{4}{3} \Rightarrow 4: 3$
Given 3p orbital In $3 \mathrm{p}$ orbital, $\mathrm{n}=3$ and $l=1$ We know, $\text { Spherical Node }=\mathrm{n}-l-1=3-1-1=1$ Non spherical Node $=l=1 Hence, one spherical Node and one non spherical Node
J and K CET-(1998)
Structure of Atom
238840
Among the first lines of Lyman, Balmer, Paschen and Brackett series in hydrogen atomic spectra, which has the highest energy?
1 Lyman
2 Balmer
3 Paschen
4 Brackett
Explanation:
Energy, $\Delta \mathrm{E}=13.6\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \mathrm{eV}$ For the first line of Lyman series, $\begin{aligned} & \mathrm{n}_1=1, \mathrm{n}_2=2 \\ & \Delta \mathrm{E}=13.6\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \mathrm{eV}=10.2 \mathrm{eV} \end{aligned}$ and energy decreases as we move on to the next series.
AP-EAMCET-1999
Structure of Atom
238842
The wavelength of spectral line emitted by hydrogen atom in the Lyman series is $\frac{16}{15 R} \mathrm{~cm}$. What is the value of $n_2$ ? (R=Rydberg constant)
1 2
2 3
3 4
4 1
Explanation:
Given, $\lambda=\frac{16}{15 \mathrm{R}} \mathrm{cm}, \mathrm{n}_2=$ ? For Lyman series- $\begin{array}{ll} & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ & \frac{15 \mathrm{R}}{16}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ \text { or } \quad \frac{15 \mathrm{R}}{16 \mathrm{R}}=1-\frac{1}{\mathrm{n}_2^2} \\ \text { or } & \frac{1}{\mathrm{n}_2^2}=1-\frac{15}{16}=\frac{1}{16} \\ \text { or } & \frac{1}{\mathrm{n}_2}=\frac{1}{4} \\ \text { or } & \mathrm{n}_2=4 \end{array}$
AP-EAMCET-2007
Structure of Atom
238843
An electronic transition in hydrogen atom results in the formation of $\mathbf{H}_\alpha$ line of hydrogen in Lyman series, the energies associated with the electron in each of the orbits involved in the transition (in kcal $\mathbf{~ m o l}^{-1}$ ) are :
1 $-313.6,-34.84$
2 $-313.6,-78.4$
3 $-78.4,-34.84$
4 $-78.4,-19.6$
Explanation:
Energy of electron in $\mathrm{n}^{\text {th }}$ orbit- $\mathrm{E}_{\mathrm{n}}=\frac{2 \pi^2 Z^2 \mathrm{e}^4}{\mathrm{n}^2 \mathrm{~h}^2} \mathrm{a}$ Putting the value of $\mathrm{h}$, e and $\pi$, we get $\begin{aligned} \mathrm{E}_{\mathrm{n}} & =\frac{-1311.8 \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{~kJ} / \mathrm{mole} \\ \mathrm{E}_{\mathrm{n}} & =\frac{-313.52 \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{kcal} / \mathrm{mole} \\ & (\because 1 \mathrm{kcal}=4.184 \mathrm{~kJ}) \end{aligned}$ For $\mathrm{H}$ atom, $\mathrm{Z}=1$ and Lyman series have the value of $\mathrm{n}_1=1, \mathrm{n}_2=2$. $\mathrm{n}_1=1, \mathrm{n}_2=2$ Since, Energy of electron in $\mathrm{n}_1$ orbit $\begin{aligned} & =\frac{-313.52 \times(1)^2}{(1)^2} \\ & =-313.6 \mathrm{kcal} / \mathrm{mole} \end{aligned}$ and energy of electron in $\mathrm{n}_2$ orbit $\begin{aligned} & =\frac{-313.52 \times(1)^2}{(2)^2} \\ & =-78.38 \mathrm{kcal} / \mathrm{mole} \end{aligned}$
AP-EAMCET-2008
Structure of Atom
238844
The ratio of the highest to the lowest wavelength of Lyman series is
1 $4: 3$
2 $9: 8$
3 $27: 5$
4 $16: 5$
Explanation:
The equation of Lyman series wavelength for hydrogen atom is $\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}}\left(1-\frac{1}{\mathrm{n}^2}\right)$ Where, $\lambda=$ Lyman series wavelength for hydrogen atom $mathrm{R}_{\mathrm{H}}=\text { Rydberg Constant }$ For lowest wavelength, $\mathrm{n}=\infty$ $\therefore \frac{1}{\lambda_L}=\mathrm{R}_{\mathrm{H}}\left(1-\frac{1}{\infty^2}\right)$ or $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}_{\mathrm{H}}$ For highest wavelength, $\mathrm{n}=2$ $\therefore \frac{1}{\lambda_{\mathrm{H}}}=\mathrm{R}_{\mathrm{H}}\left(1-\frac{1}{4}\right)=\frac{3 \mathrm{R}_{\mathrm{H}}}{4}$ From equation (1) and (2) $\frac{\lambda_{\mathrm{H}}}{\lambda_{\mathrm{L}}}=\frac{4}{3} \Rightarrow 4: 3$
Given 3p orbital In $3 \mathrm{p}$ orbital, $\mathrm{n}=3$ and $l=1$ We know, $\text { Spherical Node }=\mathrm{n}-l-1=3-1-1=1$ Non spherical Node $=l=1 Hence, one spherical Node and one non spherical Node
J and K CET-(1998)
Structure of Atom
238840
Among the first lines of Lyman, Balmer, Paschen and Brackett series in hydrogen atomic spectra, which has the highest energy?
1 Lyman
2 Balmer
3 Paschen
4 Brackett
Explanation:
Energy, $\Delta \mathrm{E}=13.6\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \mathrm{eV}$ For the first line of Lyman series, $\begin{aligned} & \mathrm{n}_1=1, \mathrm{n}_2=2 \\ & \Delta \mathrm{E}=13.6\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \mathrm{eV}=10.2 \mathrm{eV} \end{aligned}$ and energy decreases as we move on to the next series.
AP-EAMCET-1999
Structure of Atom
238842
The wavelength of spectral line emitted by hydrogen atom in the Lyman series is $\frac{16}{15 R} \mathrm{~cm}$. What is the value of $n_2$ ? (R=Rydberg constant)
1 2
2 3
3 4
4 1
Explanation:
Given, $\lambda=\frac{16}{15 \mathrm{R}} \mathrm{cm}, \mathrm{n}_2=$ ? For Lyman series- $\begin{array}{ll} & \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ & \frac{15 \mathrm{R}}{16}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ \text { or } \quad \frac{15 \mathrm{R}}{16 \mathrm{R}}=1-\frac{1}{\mathrm{n}_2^2} \\ \text { or } & \frac{1}{\mathrm{n}_2^2}=1-\frac{15}{16}=\frac{1}{16} \\ \text { or } & \frac{1}{\mathrm{n}_2}=\frac{1}{4} \\ \text { or } & \mathrm{n}_2=4 \end{array}$
AP-EAMCET-2007
Structure of Atom
238843
An electronic transition in hydrogen atom results in the formation of $\mathbf{H}_\alpha$ line of hydrogen in Lyman series, the energies associated with the electron in each of the orbits involved in the transition (in kcal $\mathbf{~ m o l}^{-1}$ ) are :
1 $-313.6,-34.84$
2 $-313.6,-78.4$
3 $-78.4,-34.84$
4 $-78.4,-19.6$
Explanation:
Energy of electron in $\mathrm{n}^{\text {th }}$ orbit- $\mathrm{E}_{\mathrm{n}}=\frac{2 \pi^2 Z^2 \mathrm{e}^4}{\mathrm{n}^2 \mathrm{~h}^2} \mathrm{a}$ Putting the value of $\mathrm{h}$, e and $\pi$, we get $\begin{aligned} \mathrm{E}_{\mathrm{n}} & =\frac{-1311.8 \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{~kJ} / \mathrm{mole} \\ \mathrm{E}_{\mathrm{n}} & =\frac{-313.52 \mathrm{Z}^2}{\mathrm{n}^2} \mathrm{kcal} / \mathrm{mole} \\ & (\because 1 \mathrm{kcal}=4.184 \mathrm{~kJ}) \end{aligned}$ For $\mathrm{H}$ atom, $\mathrm{Z}=1$ and Lyman series have the value of $\mathrm{n}_1=1, \mathrm{n}_2=2$. $\mathrm{n}_1=1, \mathrm{n}_2=2$ Since, Energy of electron in $\mathrm{n}_1$ orbit $\begin{aligned} & =\frac{-313.52 \times(1)^2}{(1)^2} \\ & =-313.6 \mathrm{kcal} / \mathrm{mole} \end{aligned}$ and energy of electron in $\mathrm{n}_2$ orbit $\begin{aligned} & =\frac{-313.52 \times(1)^2}{(2)^2} \\ & =-78.38 \mathrm{kcal} / \mathrm{mole} \end{aligned}$
AP-EAMCET-2008
Structure of Atom
238844
The ratio of the highest to the lowest wavelength of Lyman series is
1 $4: 3$
2 $9: 8$
3 $27: 5$
4 $16: 5$
Explanation:
The equation of Lyman series wavelength for hydrogen atom is $\frac{1}{\lambda}=\mathrm{R}_{\mathrm{H}}\left(1-\frac{1}{\mathrm{n}^2}\right)$ Where, $\lambda=$ Lyman series wavelength for hydrogen atom $mathrm{R}_{\mathrm{H}}=\text { Rydberg Constant }$ For lowest wavelength, $\mathrm{n}=\infty$ $\therefore \frac{1}{\lambda_L}=\mathrm{R}_{\mathrm{H}}\left(1-\frac{1}{\infty^2}\right)$ or $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}_{\mathrm{H}}$ For highest wavelength, $\mathrm{n}=2$ $\therefore \frac{1}{\lambda_{\mathrm{H}}}=\mathrm{R}_{\mathrm{H}}\left(1-\frac{1}{4}\right)=\frac{3 \mathrm{R}_{\mathrm{H}}}{4}$ From equation (1) and (2) $\frac{\lambda_{\mathrm{H}}}{\lambda_{\mathrm{L}}}=\frac{4}{3} \Rightarrow 4: 3$
