238848
Which of the following electron transitions in the $H$-atom will release the largest amount of energy?
1 For, $n=3$ to $n=2$ $\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36}$
2 For, $n=2$ to $n=1$ $\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\left(\frac{1}{1}-\frac{1}{4}\right)=\frac{3}{4}$
3 For, $\mathrm{n}=5$ to $\mathrm{n}=2$ $\left(\frac{1}{2^2}-\frac{1}{5^2}\right)=\left(\frac{1}{4}-\frac{1}{2^5}\right)=\frac{21}{100}$
4 For, $\mathrm{n}=6$ to $\mathrm{n}=2$ $\begin{aligned} & \left(\frac{1}{2^2}-\frac{1}{6^2}\right)=\left(\frac{1}{4}-\frac{1}{36}\right)=\frac{8}{36} \\ & \frac{3}{4}>\frac{21}{100}>\frac{8}{36}>\frac{5}{36} \end{aligned}$ Hence, largest amount of energy is released when electron jumps from $\mathrm{n}=2$ to $\mathrm{n}=1$.
Explanation:
As we know, $\mathrm{E}=\frac{\mathrm{hc}}{\lambda} \text { and } \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]$ So, we can say- $\mathrm{E} \propto\left\{\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right\}$ Energy released when electron jumps from (a.) For, $n=3$ to $n=2$ $\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36}$ (b.) For, $n=2$ to $n=1$ $\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\left(\frac{1}{1}-\frac{1}{4}\right)=\frac{3}{4}$ (c.) For, $\mathrm{n}=5$ to $\mathrm{n}=2$ $\left(\frac{1}{2^2}-\frac{1}{5^2}\right)=\left(\frac{1}{4}-\frac{1}{2^5}\right)=\frac{21}{100}$ (d.) For, $\mathrm{n}=6$ to $\mathrm{n}=2$ $\begin{aligned} & \left(\frac{1}{2^2}-\frac{1}{6^2}\right)=\left(\frac{1}{4}-\frac{1}{36}\right)=\frac{8}{36} \\ & \frac{3}{4}>\frac{21}{100}>\frac{8}{36}>\frac{5}{36} \end{aligned}$ Hence, largest amount of energy is released when electron jumps from $\mathrm{n}=2$ to $\mathrm{n}=1$.
COMEDK-2012
Structure of Atom
238846
Which among the following represents Schrodinger wave equation? #[Qdiff: Hard, QCat: Numerical Based, examname: GUJCET-2008 , AP EAMCET (Engg.) 18.9.2020 Shift-I]#
The Schrondinger wave equation is $\frac{\mathrm{d}^2 \psi}{\mathrm{dx}^2}+\frac{\mathrm{d}^2 \psi}{\mathrm{dy}^2}+\frac{\mathrm{d}^2 \psi}{\mathrm{dz}^2}+\frac{8 \pi^2 \mathrm{~m}}{\mathrm{~h}^2}(\mathrm{E}-\mathrm{v}) \psi=0$ Where, $\psi=$ Wave function $\mathrm{m}=$ Mass of electron, $\mathrm{h}=$ Planck's constant $\mathrm{E}=$ Total energy of electron $\mathrm{V}=$ Potential energy of electron
Structure of Atom
238847
The spectrum of Helium is expected to be similar to that of #[Qdiff: Hard, QCat: Numerical Based, examname: AP EAPCET 19-08-2021 Shift-I , NEET-1998]#
1 $\mathrm{Li}^{+}$
2 $\mathrm{H}$
3 $\mathrm{Na}$
4 $\mathrm{He}^{+}$
Explanation:
The spectrum of any element depends upon the no. of electron in outer shell. Electronic configuration of $\mathrm{He}={ }_1 \mathrm{~s}^2$ Reason:- $\mathrm{Li}^{+}$has the same electronic configuration like as $\mathrm{He}$ atom i.e. $\mathrm{Li}^{+}=1 \mathrm{~s}^2, 2 \mathrm{~s}^1$ (Removing of one electron form outer shell) $\mathrm{Li}^{+}=1 \mathrm{~s}^2$ i.e $\mathrm{Li}^{+}=2$
Structure of Atom
238851
What are the values of $n_1$ and $n_2$ respectively for $H_\beta$ line in the Lyman series of hydrogen atomic spectrum?
1 3 and 5
2 2 and 3
3 1 and 3
4 2 and 4
Explanation:
The Lyman series of lines are the lines in the hydrogen spectrum which appear in the ultraviolet region. The value of $n_1$ and $n_2$ is given as follows- $\begin{aligned} & \mathrm{n}_1=1 \\ & \mathrm{n}_2=2,3 \ldots \end{aligned}$
238848
Which of the following electron transitions in the $H$-atom will release the largest amount of energy?
1 For, $n=3$ to $n=2$ $\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36}$
2 For, $n=2$ to $n=1$ $\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\left(\frac{1}{1}-\frac{1}{4}\right)=\frac{3}{4}$
3 For, $\mathrm{n}=5$ to $\mathrm{n}=2$ $\left(\frac{1}{2^2}-\frac{1}{5^2}\right)=\left(\frac{1}{4}-\frac{1}{2^5}\right)=\frac{21}{100}$
4 For, $\mathrm{n}=6$ to $\mathrm{n}=2$ $\begin{aligned} & \left(\frac{1}{2^2}-\frac{1}{6^2}\right)=\left(\frac{1}{4}-\frac{1}{36}\right)=\frac{8}{36} \\ & \frac{3}{4}>\frac{21}{100}>\frac{8}{36}>\frac{5}{36} \end{aligned}$ Hence, largest amount of energy is released when electron jumps from $\mathrm{n}=2$ to $\mathrm{n}=1$.
Explanation:
As we know, $\mathrm{E}=\frac{\mathrm{hc}}{\lambda} \text { and } \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]$ So, we can say- $\mathrm{E} \propto\left\{\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right\}$ Energy released when electron jumps from (a.) For, $n=3$ to $n=2$ $\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36}$ (b.) For, $n=2$ to $n=1$ $\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\left(\frac{1}{1}-\frac{1}{4}\right)=\frac{3}{4}$ (c.) For, $\mathrm{n}=5$ to $\mathrm{n}=2$ $\left(\frac{1}{2^2}-\frac{1}{5^2}\right)=\left(\frac{1}{4}-\frac{1}{2^5}\right)=\frac{21}{100}$ (d.) For, $\mathrm{n}=6$ to $\mathrm{n}=2$ $\begin{aligned} & \left(\frac{1}{2^2}-\frac{1}{6^2}\right)=\left(\frac{1}{4}-\frac{1}{36}\right)=\frac{8}{36} \\ & \frac{3}{4}>\frac{21}{100}>\frac{8}{36}>\frac{5}{36} \end{aligned}$ Hence, largest amount of energy is released when electron jumps from $\mathrm{n}=2$ to $\mathrm{n}=1$.
COMEDK-2012
Structure of Atom
238846
Which among the following represents Schrodinger wave equation? #[Qdiff: Hard, QCat: Numerical Based, examname: GUJCET-2008 , AP EAMCET (Engg.) 18.9.2020 Shift-I]#
The Schrondinger wave equation is $\frac{\mathrm{d}^2 \psi}{\mathrm{dx}^2}+\frac{\mathrm{d}^2 \psi}{\mathrm{dy}^2}+\frac{\mathrm{d}^2 \psi}{\mathrm{dz}^2}+\frac{8 \pi^2 \mathrm{~m}}{\mathrm{~h}^2}(\mathrm{E}-\mathrm{v}) \psi=0$ Where, $\psi=$ Wave function $\mathrm{m}=$ Mass of electron, $\mathrm{h}=$ Planck's constant $\mathrm{E}=$ Total energy of electron $\mathrm{V}=$ Potential energy of electron
Structure of Atom
238847
The spectrum of Helium is expected to be similar to that of #[Qdiff: Hard, QCat: Numerical Based, examname: AP EAPCET 19-08-2021 Shift-I , NEET-1998]#
1 $\mathrm{Li}^{+}$
2 $\mathrm{H}$
3 $\mathrm{Na}$
4 $\mathrm{He}^{+}$
Explanation:
The spectrum of any element depends upon the no. of electron in outer shell. Electronic configuration of $\mathrm{He}={ }_1 \mathrm{~s}^2$ Reason:- $\mathrm{Li}^{+}$has the same electronic configuration like as $\mathrm{He}$ atom i.e. $\mathrm{Li}^{+}=1 \mathrm{~s}^2, 2 \mathrm{~s}^1$ (Removing of one electron form outer shell) $\mathrm{Li}^{+}=1 \mathrm{~s}^2$ i.e $\mathrm{Li}^{+}=2$
Structure of Atom
238851
What are the values of $n_1$ and $n_2$ respectively for $H_\beta$ line in the Lyman series of hydrogen atomic spectrum?
1 3 and 5
2 2 and 3
3 1 and 3
4 2 and 4
Explanation:
The Lyman series of lines are the lines in the hydrogen spectrum which appear in the ultraviolet region. The value of $n_1$ and $n_2$ is given as follows- $\begin{aligned} & \mathrm{n}_1=1 \\ & \mathrm{n}_2=2,3 \ldots \end{aligned}$
238848
Which of the following electron transitions in the $H$-atom will release the largest amount of energy?
1 For, $n=3$ to $n=2$ $\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36}$
2 For, $n=2$ to $n=1$ $\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\left(\frac{1}{1}-\frac{1}{4}\right)=\frac{3}{4}$
3 For, $\mathrm{n}=5$ to $\mathrm{n}=2$ $\left(\frac{1}{2^2}-\frac{1}{5^2}\right)=\left(\frac{1}{4}-\frac{1}{2^5}\right)=\frac{21}{100}$
4 For, $\mathrm{n}=6$ to $\mathrm{n}=2$ $\begin{aligned} & \left(\frac{1}{2^2}-\frac{1}{6^2}\right)=\left(\frac{1}{4}-\frac{1}{36}\right)=\frac{8}{36} \\ & \frac{3}{4}>\frac{21}{100}>\frac{8}{36}>\frac{5}{36} \end{aligned}$ Hence, largest amount of energy is released when electron jumps from $\mathrm{n}=2$ to $\mathrm{n}=1$.
Explanation:
As we know, $\mathrm{E}=\frac{\mathrm{hc}}{\lambda} \text { and } \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]$ So, we can say- $\mathrm{E} \propto\left\{\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right\}$ Energy released when electron jumps from (a.) For, $n=3$ to $n=2$ $\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36}$ (b.) For, $n=2$ to $n=1$ $\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\left(\frac{1}{1}-\frac{1}{4}\right)=\frac{3}{4}$ (c.) For, $\mathrm{n}=5$ to $\mathrm{n}=2$ $\left(\frac{1}{2^2}-\frac{1}{5^2}\right)=\left(\frac{1}{4}-\frac{1}{2^5}\right)=\frac{21}{100}$ (d.) For, $\mathrm{n}=6$ to $\mathrm{n}=2$ $\begin{aligned} & \left(\frac{1}{2^2}-\frac{1}{6^2}\right)=\left(\frac{1}{4}-\frac{1}{36}\right)=\frac{8}{36} \\ & \frac{3}{4}>\frac{21}{100}>\frac{8}{36}>\frac{5}{36} \end{aligned}$ Hence, largest amount of energy is released when electron jumps from $\mathrm{n}=2$ to $\mathrm{n}=1$.
COMEDK-2012
Structure of Atom
238846
Which among the following represents Schrodinger wave equation? #[Qdiff: Hard, QCat: Numerical Based, examname: GUJCET-2008 , AP EAMCET (Engg.) 18.9.2020 Shift-I]#
The Schrondinger wave equation is $\frac{\mathrm{d}^2 \psi}{\mathrm{dx}^2}+\frac{\mathrm{d}^2 \psi}{\mathrm{dy}^2}+\frac{\mathrm{d}^2 \psi}{\mathrm{dz}^2}+\frac{8 \pi^2 \mathrm{~m}}{\mathrm{~h}^2}(\mathrm{E}-\mathrm{v}) \psi=0$ Where, $\psi=$ Wave function $\mathrm{m}=$ Mass of electron, $\mathrm{h}=$ Planck's constant $\mathrm{E}=$ Total energy of electron $\mathrm{V}=$ Potential energy of electron
Structure of Atom
238847
The spectrum of Helium is expected to be similar to that of #[Qdiff: Hard, QCat: Numerical Based, examname: AP EAPCET 19-08-2021 Shift-I , NEET-1998]#
1 $\mathrm{Li}^{+}$
2 $\mathrm{H}$
3 $\mathrm{Na}$
4 $\mathrm{He}^{+}$
Explanation:
The spectrum of any element depends upon the no. of electron in outer shell. Electronic configuration of $\mathrm{He}={ }_1 \mathrm{~s}^2$ Reason:- $\mathrm{Li}^{+}$has the same electronic configuration like as $\mathrm{He}$ atom i.e. $\mathrm{Li}^{+}=1 \mathrm{~s}^2, 2 \mathrm{~s}^1$ (Removing of one electron form outer shell) $\mathrm{Li}^{+}=1 \mathrm{~s}^2$ i.e $\mathrm{Li}^{+}=2$
Structure of Atom
238851
What are the values of $n_1$ and $n_2$ respectively for $H_\beta$ line in the Lyman series of hydrogen atomic spectrum?
1 3 and 5
2 2 and 3
3 1 and 3
4 2 and 4
Explanation:
The Lyman series of lines are the lines in the hydrogen spectrum which appear in the ultraviolet region. The value of $n_1$ and $n_2$ is given as follows- $\begin{aligned} & \mathrm{n}_1=1 \\ & \mathrm{n}_2=2,3 \ldots \end{aligned}$
238848
Which of the following electron transitions in the $H$-atom will release the largest amount of energy?
1 For, $n=3$ to $n=2$ $\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36}$
2 For, $n=2$ to $n=1$ $\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\left(\frac{1}{1}-\frac{1}{4}\right)=\frac{3}{4}$
3 For, $\mathrm{n}=5$ to $\mathrm{n}=2$ $\left(\frac{1}{2^2}-\frac{1}{5^2}\right)=\left(\frac{1}{4}-\frac{1}{2^5}\right)=\frac{21}{100}$
4 For, $\mathrm{n}=6$ to $\mathrm{n}=2$ $\begin{aligned} & \left(\frac{1}{2^2}-\frac{1}{6^2}\right)=\left(\frac{1}{4}-\frac{1}{36}\right)=\frac{8}{36} \\ & \frac{3}{4}>\frac{21}{100}>\frac{8}{36}>\frac{5}{36} \end{aligned}$ Hence, largest amount of energy is released when electron jumps from $\mathrm{n}=2$ to $\mathrm{n}=1$.
Explanation:
As we know, $\mathrm{E}=\frac{\mathrm{hc}}{\lambda} \text { and } \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]$ So, we can say- $\mathrm{E} \propto\left\{\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right\}$ Energy released when electron jumps from (a.) For, $n=3$ to $n=2$ $\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36}$ (b.) For, $n=2$ to $n=1$ $\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\left(\frac{1}{1}-\frac{1}{4}\right)=\frac{3}{4}$ (c.) For, $\mathrm{n}=5$ to $\mathrm{n}=2$ $\left(\frac{1}{2^2}-\frac{1}{5^2}\right)=\left(\frac{1}{4}-\frac{1}{2^5}\right)=\frac{21}{100}$ (d.) For, $\mathrm{n}=6$ to $\mathrm{n}=2$ $\begin{aligned} & \left(\frac{1}{2^2}-\frac{1}{6^2}\right)=\left(\frac{1}{4}-\frac{1}{36}\right)=\frac{8}{36} \\ & \frac{3}{4}>\frac{21}{100}>\frac{8}{36}>\frac{5}{36} \end{aligned}$ Hence, largest amount of energy is released when electron jumps from $\mathrm{n}=2$ to $\mathrm{n}=1$.
COMEDK-2012
Structure of Atom
238846
Which among the following represents Schrodinger wave equation? #[Qdiff: Hard, QCat: Numerical Based, examname: GUJCET-2008 , AP EAMCET (Engg.) 18.9.2020 Shift-I]#
The Schrondinger wave equation is $\frac{\mathrm{d}^2 \psi}{\mathrm{dx}^2}+\frac{\mathrm{d}^2 \psi}{\mathrm{dy}^2}+\frac{\mathrm{d}^2 \psi}{\mathrm{dz}^2}+\frac{8 \pi^2 \mathrm{~m}}{\mathrm{~h}^2}(\mathrm{E}-\mathrm{v}) \psi=0$ Where, $\psi=$ Wave function $\mathrm{m}=$ Mass of electron, $\mathrm{h}=$ Planck's constant $\mathrm{E}=$ Total energy of electron $\mathrm{V}=$ Potential energy of electron
Structure of Atom
238847
The spectrum of Helium is expected to be similar to that of #[Qdiff: Hard, QCat: Numerical Based, examname: AP EAPCET 19-08-2021 Shift-I , NEET-1998]#
1 $\mathrm{Li}^{+}$
2 $\mathrm{H}$
3 $\mathrm{Na}$
4 $\mathrm{He}^{+}$
Explanation:
The spectrum of any element depends upon the no. of electron in outer shell. Electronic configuration of $\mathrm{He}={ }_1 \mathrm{~s}^2$ Reason:- $\mathrm{Li}^{+}$has the same electronic configuration like as $\mathrm{He}$ atom i.e. $\mathrm{Li}^{+}=1 \mathrm{~s}^2, 2 \mathrm{~s}^1$ (Removing of one electron form outer shell) $\mathrm{Li}^{+}=1 \mathrm{~s}^2$ i.e $\mathrm{Li}^{+}=2$
Structure of Atom
238851
What are the values of $n_1$ and $n_2$ respectively for $H_\beta$ line in the Lyman series of hydrogen atomic spectrum?
1 3 and 5
2 2 and 3
3 1 and 3
4 2 and 4
Explanation:
The Lyman series of lines are the lines in the hydrogen spectrum which appear in the ultraviolet region. The value of $n_1$ and $n_2$ is given as follows- $\begin{aligned} & \mathrm{n}_1=1 \\ & \mathrm{n}_2=2,3 \ldots \end{aligned}$
