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PHXII12:ATOMS

356485 The angular momentum of an electron in a hydrogen atom is proportional to (where $r$ is the radius of the orbit of electron)

1

\frac{1}{\sqrt{r}}

2

\frac{1}{r}

3

\sqrt{r}

4

r

Explanation:

Angular momentum of an electron in a hydrogen atom.
$L = \frac{n h}{2 π}$
$\Rightarrow L \propto n \propto \sqrt{r} (∵ r \propto n^{2} \Rightarrow n \propto \sqrt{r}) (1)$

JEE - 2024

PHXII12:ATOMS

356486 The energy of hydrogen atom in ground state is $13.6 e V$ The ionisation energy of singly ionised helium $H e^{2 +}$ will be

1

13.6 e V

2

3.4 e V

3

27.2 e V

4

54.4 e V

Explanation:

Energy of ionised $H e^{2 +}$ ,
$E = (E_{0}) Z^{2} = (13.6) (4) = 54.4 e V$

AIIMS - 2009

PHXII12:ATOMS

356487 Read the Statement - A and Statement - B carefully to mark correct options given below
Statement A :
Atoms are electrically neutral as they contain equal number of positive and negative charges.
Statement B :
Atoms of each element are stable and emit their characteristic spectrum.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

NEET - 2024

PHXII12:ATOMS

356488 If the speed of an electron of hydrogen atom in the ground state is $2.2 \times 10^{6} m / s$ , then its speed in the third excited state will be

1

5.5 \times 10^{6} m / s

2

5.5 \times 10^{5} m / s

3

8.8 \times 10^{5} m / s

4

6.8 \times 10^{6} m / s

Explanation:

Speed of electron of hydrogen atom in ground state,
$v_{1} = 2.2 \times 10^{6} m / s$
Since, $v \propto \frac{1}{n}$
$∴ \frac{v_{1}}{v_{2}} = \frac{n_{1}}{n_{2}} = \frac{2.2 \times 10^{6}}{v_{2}}$
[ for third excited state, $n_{2} = 4$ ]
$\Rightarrow v_{2} = \frac{2.2 \times 10^{6}}{4} = 0.55 \times 10^{6} m / s$
$= 5.5 \times 10^{5} m / s$

MHTCET - 2019

PHXII12:ATOMS

356489 The electron in hydrogen atom jumps from $1^{s t}$ orbit to the $5^{t h}$ orbit. Change in angular momentum of electron is

1

\frac{2 h}{π}

2

\frac{3 h}{2 π}

3

\frac{h}{π}

4

\frac{h}{2 π}

Explanation:

Angular momentum of electron is
$L = \frac{n h}{2 π}$
change in $L$ is $Δ L = \frac{5 h}{2 π} - \frac{h}{2 π} = \frac{2 h}{π}$

PHXII12:ATOMS

356485 The angular momentum of an electron in a hydrogen atom is proportional to (where $r$ is the radius of the orbit of electron)

1

\frac{1}{\sqrt{r}}

2

\frac{1}{r}

3

\sqrt{r}

4

r

Explanation:

Angular momentum of an electron in a hydrogen atom.
$L = \frac{n h}{2 π}$
$\Rightarrow L \propto n \propto \sqrt{r} (∵ r \propto n^{2} \Rightarrow n \propto \sqrt{r}) (1)$

JEE - 2024

PHXII12:ATOMS

356486 The energy of hydrogen atom in ground state is $13.6 e V$ The ionisation energy of singly ionised helium $H e^{2 +}$ will be

1

13.6 e V

2

3.4 e V

3

27.2 e V

4

54.4 e V

Explanation:

Energy of ionised $H e^{2 +}$ ,
$E = (E_{0}) Z^{2} = (13.6) (4) = 54.4 e V$

AIIMS - 2009

PHXII12:ATOMS

356487 Read the Statement - A and Statement - B carefully to mark correct options given below
Statement A :
Atoms are electrically neutral as they contain equal number of positive and negative charges.
Statement B :
Atoms of each element are stable and emit their characteristic spectrum.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

NEET - 2024

PHXII12:ATOMS

356488 If the speed of an electron of hydrogen atom in the ground state is $2.2 \times 10^{6} m / s$ , then its speed in the third excited state will be

1

5.5 \times 10^{6} m / s

2

5.5 \times 10^{5} m / s

3

8.8 \times 10^{5} m / s

4

6.8 \times 10^{6} m / s

Explanation:

Speed of electron of hydrogen atom in ground state,
$v_{1} = 2.2 \times 10^{6} m / s$
Since, $v \propto \frac{1}{n}$
$∴ \frac{v_{1}}{v_{2}} = \frac{n_{1}}{n_{2}} = \frac{2.2 \times 10^{6}}{v_{2}}$
[ for third excited state, $n_{2} = 4$ ]
$\Rightarrow v_{2} = \frac{2.2 \times 10^{6}}{4} = 0.55 \times 10^{6} m / s$
$= 5.5 \times 10^{5} m / s$

MHTCET - 2019

PHXII12:ATOMS

356489 The electron in hydrogen atom jumps from $1^{s t}$ orbit to the $5^{t h}$ orbit. Change in angular momentum of electron is

1

\frac{2 h}{π}

2

\frac{3 h}{2 π}

3

\frac{h}{π}

4

\frac{h}{2 π}

Explanation:

Angular momentum of electron is
$L = \frac{n h}{2 π}$
change in $L$ is $Δ L = \frac{5 h}{2 π} - \frac{h}{2 π} = \frac{2 h}{π}$

PHXII12:ATOMS

356485 The angular momentum of an electron in a hydrogen atom is proportional to (where $r$ is the radius of the orbit of electron)

1

\frac{1}{\sqrt{r}}

2

\frac{1}{r}

3

\sqrt{r}

4

r

Explanation:

Angular momentum of an electron in a hydrogen atom.
$L = \frac{n h}{2 π}$
$\Rightarrow L \propto n \propto \sqrt{r} (∵ r \propto n^{2} \Rightarrow n \propto \sqrt{r}) (1)$

JEE - 2024

PHXII12:ATOMS

356486 The energy of hydrogen atom in ground state is $13.6 e V$ The ionisation energy of singly ionised helium $H e^{2 +}$ will be

1

13.6 e V

2

3.4 e V

3

27.2 e V

4

54.4 e V

Explanation:

Energy of ionised $H e^{2 +}$ ,
$E = (E_{0}) Z^{2} = (13.6) (4) = 54.4 e V$

AIIMS - 2009

PHXII12:ATOMS

356487 Read the Statement - A and Statement - B carefully to mark correct options given below
Statement A :
Atoms are electrically neutral as they contain equal number of positive and negative charges.
Statement B :
Atoms of each element are stable and emit their characteristic spectrum.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

NEET - 2024

PHXII12:ATOMS

356488 If the speed of an electron of hydrogen atom in the ground state is $2.2 \times 10^{6} m / s$ , then its speed in the third excited state will be

1

5.5 \times 10^{6} m / s

2

5.5 \times 10^{5} m / s

3

8.8 \times 10^{5} m / s

4

6.8 \times 10^{6} m / s

Explanation:

Speed of electron of hydrogen atom in ground state,
$v_{1} = 2.2 \times 10^{6} m / s$
Since, $v \propto \frac{1}{n}$
$∴ \frac{v_{1}}{v_{2}} = \frac{n_{1}}{n_{2}} = \frac{2.2 \times 10^{6}}{v_{2}}$
[ for third excited state, $n_{2} = 4$ ]
$\Rightarrow v_{2} = \frac{2.2 \times 10^{6}}{4} = 0.55 \times 10^{6} m / s$
$= 5.5 \times 10^{5} m / s$

MHTCET - 2019

PHXII12:ATOMS

356489 The electron in hydrogen atom jumps from $1^{s t}$ orbit to the $5^{t h}$ orbit. Change in angular momentum of electron is

1

\frac{2 h}{π}

2

\frac{3 h}{2 π}

3

\frac{h}{π}

4

\frac{h}{2 π}

Explanation:

Angular momentum of electron is
$L = \frac{n h}{2 π}$
change in $L$ is $Δ L = \frac{5 h}{2 π} - \frac{h}{2 π} = \frac{2 h}{π}$

PHXII12:ATOMS

356485 The angular momentum of an electron in a hydrogen atom is proportional to (where $r$ is the radius of the orbit of electron)

1

\frac{1}{\sqrt{r}}

2

\frac{1}{r}

3

\sqrt{r}

4

r

Explanation:

Angular momentum of an electron in a hydrogen atom.
$L = \frac{n h}{2 π}$
$\Rightarrow L \propto n \propto \sqrt{r} (∵ r \propto n^{2} \Rightarrow n \propto \sqrt{r}) (1)$

JEE - 2024

PHXII12:ATOMS

356486 The energy of hydrogen atom in ground state is $13.6 e V$ The ionisation energy of singly ionised helium $H e^{2 +}$ will be

1

13.6 e V

2

3.4 e V

3

27.2 e V

4

54.4 e V

Explanation:

Energy of ionised $H e^{2 +}$ ,
$E = (E_{0}) Z^{2} = (13.6) (4) = 54.4 e V$

AIIMS - 2009

PHXII12:ATOMS

356487 Read the Statement - A and Statement - B carefully to mark correct options given below
Statement A :
Atoms are electrically neutral as they contain equal number of positive and negative charges.
Statement B :
Atoms of each element are stable and emit their characteristic spectrum.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

NEET - 2024

PHXII12:ATOMS

356488 If the speed of an electron of hydrogen atom in the ground state is $2.2 \times 10^{6} m / s$ , then its speed in the third excited state will be

1

5.5 \times 10^{6} m / s

2

5.5 \times 10^{5} m / s

3

8.8 \times 10^{5} m / s

4

6.8 \times 10^{6} m / s

Explanation:

Speed of electron of hydrogen atom in ground state,
$v_{1} = 2.2 \times 10^{6} m / s$
Since, $v \propto \frac{1}{n}$
$∴ \frac{v_{1}}{v_{2}} = \frac{n_{1}}{n_{2}} = \frac{2.2 \times 10^{6}}{v_{2}}$
[ for third excited state, $n_{2} = 4$ ]
$\Rightarrow v_{2} = \frac{2.2 \times 10^{6}}{4} = 0.55 \times 10^{6} m / s$
$= 5.5 \times 10^{5} m / s$

MHTCET - 2019

PHXII12:ATOMS

356489 The electron in hydrogen atom jumps from $1^{s t}$ orbit to the $5^{t h}$ orbit. Change in angular momentum of electron is

1

\frac{2 h}{π}

2

\frac{3 h}{2 π}

3

\frac{h}{π}

4

\frac{h}{2 π}

Explanation:

Angular momentum of electron is
$L = \frac{n h}{2 π}$
change in $L$ is $Δ L = \frac{5 h}{2 π} - \frac{h}{2 π} = \frac{2 h}{π}$

PHXII12:ATOMS

356485 The angular momentum of an electron in a hydrogen atom is proportional to (where $r$ is the radius of the orbit of electron)

1

\frac{1}{\sqrt{r}}

2

\frac{1}{r}

3

\sqrt{r}

4

r

Explanation:

Angular momentum of an electron in a hydrogen atom.
$L = \frac{n h}{2 π}$
$\Rightarrow L \propto n \propto \sqrt{r} (∵ r \propto n^{2} \Rightarrow n \propto \sqrt{r}) (1)$

JEE - 2024

PHXII12:ATOMS

356486 The energy of hydrogen atom in ground state is $13.6 e V$ The ionisation energy of singly ionised helium $H e^{2 +}$ will be

1

13.6 e V

2

3.4 e V

3

27.2 e V

4

54.4 e V

Explanation:

Energy of ionised $H e^{2 +}$ ,
$E = (E_{0}) Z^{2} = (13.6) (4) = 54.4 e V$

AIIMS - 2009

PHXII12:ATOMS

356487 Read the Statement - A and Statement - B carefully to mark correct options given below
Statement A :
Atoms are electrically neutral as they contain equal number of positive and negative charges.
Statement B :
Atoms of each element are stable and emit their characteristic spectrum.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

NEET - 2024

PHXII12:ATOMS

356488 If the speed of an electron of hydrogen atom in the ground state is $2.2 \times 10^{6} m / s$ , then its speed in the third excited state will be

1

5.5 \times 10^{6} m / s

2

5.5 \times 10^{5} m / s

3

8.8 \times 10^{5} m / s

4

6.8 \times 10^{6} m / s

Explanation:

Speed of electron of hydrogen atom in ground state,
$v_{1} = 2.2 \times 10^{6} m / s$
Since, $v \propto \frac{1}{n}$
$∴ \frac{v_{1}}{v_{2}} = \frac{n_{1}}{n_{2}} = \frac{2.2 \times 10^{6}}{v_{2}}$
[ for third excited state, $n_{2} = 4$ ]
$\Rightarrow v_{2} = \frac{2.2 \times 10^{6}}{4} = 0.55 \times 10^{6} m / s$
$= 5.5 \times 10^{5} m / s$

MHTCET - 2019

PHXII12:ATOMS

356489 The electron in hydrogen atom jumps from $1^{s t}$ orbit to the $5^{t h}$ orbit. Change in angular momentum of electron is

1

\frac{2 h}{π}

2

\frac{3 h}{2 π}

3

\frac{h}{π}

4

\frac{h}{2 π}

Explanation:

Angular momentum of electron is
$L = \frac{n h}{2 π}$
change in $L$ is $Δ L = \frac{5 h}{2 π} - \frac{h}{2 π} = \frac{2 h}{π}$