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PHXII12:ATOMS

356391 Speed of an electron in Bohr's $7^{th}$ orbit for Hydrogen atom is $3.6 \times 10^{6} m / s$ . The corresponding speed of the electron in $3^{rd}$ orbit, in $m / s$ is

1

1.8 \times 10^{6}

2

3.6 \times 10^{6}

3

7.5 \times 10^{6}

4

8.4 \times 10^{6}

Explanation:

The speed of electron in Bohr orbit of a hydrogen - like atom is given by $V_{n} = V_{o} \frac{Z}{n}$ where $V_{o}$ is the speed in ground state.
For $H$ -atom, $Z = 1$
Here, $v_{7} = 3.6 \times 10^{6} m / s$
Therefore $\frac{v_{7}}{v_{3}} = \frac{3}{7}$
or $v_{3} = \frac{(3.6 \times 10^{6}) \times 7}{3}$
$= 8.4 \times 10^{6} m / s$

JEE - 2023

PHXII12:ATOMS

356392 An electron in Bohr's hydrogen atom has an energy of $- 3.4$ $e V$ . The angular momentum of the electron is

1

\frac{h}{π}

2

\frac{h}{2 π}

3

\frac{n h}{2 π}

(

n

is an integer)

4

\frac{2 h}{π}

Explanation:

Energy of the electron in the $n^{th}$ orbit is $E_{n} = \frac{E_{1}}{n^{2}} = \frac{- 13.6}{n^{2}} e V$
For $n = 2, E_{2} = \frac{- 13.6}{4} e V = - 3.4 e V$
Angular momentum of the electron in the $n^{th}$ orbit is
$L = m v r = \frac{n h}{2 π} \Rightarrow L = \frac{2 h}{2 π} = \frac{h}{π} (n = 2)$ .
So correct option is (1)

PHXII12:ATOMS

356393 The ground state energy of hydrogen atom is $- 13.6 e V$ . What is the potential energy?

1

0 e V

2

- 27.2 e V

3

1 e V

4

2 e V

Explanation:

$P . E = 2 (T . E) = 2 \times (- 13.6 e V) = - 27.2 e V$

PHXII12:ATOMS

356394 A hydrogen atom in ground state absorbs $10.2 e V$ of energy. The orbital angular momentum of the electron is increased by

1

3.16 \times 10^{- 34} J s

2

1.05 \times 10^{- 34} J s

3

4.22 \times 10^{- 34} J s

4

2.11 \times 10^{- 34} J s

Explanation:

Energy absorbed during electron transition from
$n = 1$ to $n = 2$
$Δ E = 13.6 \times (\frac{1}{1^{2}} - \frac{1}{2^{2}}) e V = 10.2 e V$
As the hydrogen atom in the given problem also absorbs 10.2 $e V$ , the only possible electron transition is from $n = 1$ to $n = 2$ .
$∴$ Change in angular momentum
$Δ L = \frac{(2 - 1) h}{2 π} = \frac{6.6 \times 10^{- 34}}{2 \times 3.14} = 1.05 \times 10^{- 34} J s$

KCET - 2019

PHXII12:ATOMS

356391 Speed of an electron in Bohr's $7^{th}$ orbit for Hydrogen atom is $3.6 \times 10^{6} m / s$ . The corresponding speed of the electron in $3^{rd}$ orbit, in $m / s$ is

1

1.8 \times 10^{6}

2

3.6 \times 10^{6}

3

7.5 \times 10^{6}

4

8.4 \times 10^{6}

Explanation:

The speed of electron in Bohr orbit of a hydrogen - like atom is given by $V_{n} = V_{o} \frac{Z}{n}$ where $V_{o}$ is the speed in ground state.
For $H$ -atom, $Z = 1$
Here, $v_{7} = 3.6 \times 10^{6} m / s$
Therefore $\frac{v_{7}}{v_{3}} = \frac{3}{7}$
or $v_{3} = \frac{(3.6 \times 10^{6}) \times 7}{3}$
$= 8.4 \times 10^{6} m / s$

JEE - 2023

PHXII12:ATOMS

356392 An electron in Bohr's hydrogen atom has an energy of $- 3.4$ $e V$ . The angular momentum of the electron is

1

\frac{h}{π}

2

\frac{h}{2 π}

3

\frac{n h}{2 π}

(

n

is an integer)

4

\frac{2 h}{π}

Explanation:

Energy of the electron in the $n^{th}$ orbit is $E_{n} = \frac{E_{1}}{n^{2}} = \frac{- 13.6}{n^{2}} e V$
For $n = 2, E_{2} = \frac{- 13.6}{4} e V = - 3.4 e V$
Angular momentum of the electron in the $n^{th}$ orbit is
$L = m v r = \frac{n h}{2 π} \Rightarrow L = \frac{2 h}{2 π} = \frac{h}{π} (n = 2)$ .
So correct option is (1)

PHXII12:ATOMS

356393 The ground state energy of hydrogen atom is $- 13.6 e V$ . What is the potential energy?

1

0 e V

2

- 27.2 e V

3

1 e V

4

2 e V

Explanation:

$P . E = 2 (T . E) = 2 \times (- 13.6 e V) = - 27.2 e V$

PHXII12:ATOMS

356394 A hydrogen atom in ground state absorbs $10.2 e V$ of energy. The orbital angular momentum of the electron is increased by

1

3.16 \times 10^{- 34} J s

2

1.05 \times 10^{- 34} J s

3

4.22 \times 10^{- 34} J s

4

2.11 \times 10^{- 34} J s

Explanation:

Energy absorbed during electron transition from
$n = 1$ to $n = 2$
$Δ E = 13.6 \times (\frac{1}{1^{2}} - \frac{1}{2^{2}}) e V = 10.2 e V$
As the hydrogen atom in the given problem also absorbs 10.2 $e V$ , the only possible electron transition is from $n = 1$ to $n = 2$ .
$∴$ Change in angular momentum
$Δ L = \frac{(2 - 1) h}{2 π} = \frac{6.6 \times 10^{- 34}}{2 \times 3.14} = 1.05 \times 10^{- 34} J s$

KCET - 2019

PHXII12:ATOMS

356391 Speed of an electron in Bohr's $7^{th}$ orbit for Hydrogen atom is $3.6 \times 10^{6} m / s$ . The corresponding speed of the electron in $3^{rd}$ orbit, in $m / s$ is

1

1.8 \times 10^{6}

2

3.6 \times 10^{6}

3

7.5 \times 10^{6}

4

8.4 \times 10^{6}

Explanation:

The speed of electron in Bohr orbit of a hydrogen - like atom is given by $V_{n} = V_{o} \frac{Z}{n}$ where $V_{o}$ is the speed in ground state.
For $H$ -atom, $Z = 1$
Here, $v_{7} = 3.6 \times 10^{6} m / s$
Therefore $\frac{v_{7}}{v_{3}} = \frac{3}{7}$
or $v_{3} = \frac{(3.6 \times 10^{6}) \times 7}{3}$
$= 8.4 \times 10^{6} m / s$

JEE - 2023

PHXII12:ATOMS

356392 An electron in Bohr's hydrogen atom has an energy of $- 3.4$ $e V$ . The angular momentum of the electron is

1

\frac{h}{π}

2

\frac{h}{2 π}

3

\frac{n h}{2 π}

(

n

is an integer)

4

\frac{2 h}{π}

Explanation:

Energy of the electron in the $n^{th}$ orbit is $E_{n} = \frac{E_{1}}{n^{2}} = \frac{- 13.6}{n^{2}} e V$
For $n = 2, E_{2} = \frac{- 13.6}{4} e V = - 3.4 e V$
Angular momentum of the electron in the $n^{th}$ orbit is
$L = m v r = \frac{n h}{2 π} \Rightarrow L = \frac{2 h}{2 π} = \frac{h}{π} (n = 2)$ .
So correct option is (1)

PHXII12:ATOMS

356393 The ground state energy of hydrogen atom is $- 13.6 e V$ . What is the potential energy?

1

0 e V

2

- 27.2 e V

3

1 e V

4

2 e V

Explanation:

$P . E = 2 (T . E) = 2 \times (- 13.6 e V) = - 27.2 e V$

PHXII12:ATOMS

356394 A hydrogen atom in ground state absorbs $10.2 e V$ of energy. The orbital angular momentum of the electron is increased by

1

3.16 \times 10^{- 34} J s

2

1.05 \times 10^{- 34} J s

3

4.22 \times 10^{- 34} J s

4

2.11 \times 10^{- 34} J s

Explanation:

Energy absorbed during electron transition from
$n = 1$ to $n = 2$
$Δ E = 13.6 \times (\frac{1}{1^{2}} - \frac{1}{2^{2}}) e V = 10.2 e V$
As the hydrogen atom in the given problem also absorbs 10.2 $e V$ , the only possible electron transition is from $n = 1$ to $n = 2$ .
$∴$ Change in angular momentum
$Δ L = \frac{(2 - 1) h}{2 π} = \frac{6.6 \times 10^{- 34}}{2 \times 3.14} = 1.05 \times 10^{- 34} J s$

KCET - 2019

PHXII12:ATOMS

356391 Speed of an electron in Bohr's $7^{th}$ orbit for Hydrogen atom is $3.6 \times 10^{6} m / s$ . The corresponding speed of the electron in $3^{rd}$ orbit, in $m / s$ is

1

1.8 \times 10^{6}

2

3.6 \times 10^{6}

3

7.5 \times 10^{6}

4

8.4 \times 10^{6}

Explanation:

The speed of electron in Bohr orbit of a hydrogen - like atom is given by $V_{n} = V_{o} \frac{Z}{n}$ where $V_{o}$ is the speed in ground state.
For $H$ -atom, $Z = 1$
Here, $v_{7} = 3.6 \times 10^{6} m / s$
Therefore $\frac{v_{7}}{v_{3}} = \frac{3}{7}$
or $v_{3} = \frac{(3.6 \times 10^{6}) \times 7}{3}$
$= 8.4 \times 10^{6} m / s$

JEE - 2023

PHXII12:ATOMS

356392 An electron in Bohr's hydrogen atom has an energy of $- 3.4$ $e V$ . The angular momentum of the electron is

1

\frac{h}{π}

2

\frac{h}{2 π}

3

\frac{n h}{2 π}

(

n

is an integer)

4

\frac{2 h}{π}

Explanation:

Energy of the electron in the $n^{th}$ orbit is $E_{n} = \frac{E_{1}}{n^{2}} = \frac{- 13.6}{n^{2}} e V$
For $n = 2, E_{2} = \frac{- 13.6}{4} e V = - 3.4 e V$
Angular momentum of the electron in the $n^{th}$ orbit is
$L = m v r = \frac{n h}{2 π} \Rightarrow L = \frac{2 h}{2 π} = \frac{h}{π} (n = 2)$ .
So correct option is (1)

PHXII12:ATOMS

356393 The ground state energy of hydrogen atom is $- 13.6 e V$ . What is the potential energy?

1

0 e V

2

- 27.2 e V

3

1 e V

4

2 e V

Explanation:

$P . E = 2 (T . E) = 2 \times (- 13.6 e V) = - 27.2 e V$

PHXII12:ATOMS

356394 A hydrogen atom in ground state absorbs $10.2 e V$ of energy. The orbital angular momentum of the electron is increased by

1

3.16 \times 10^{- 34} J s

2

1.05 \times 10^{- 34} J s

3

4.22 \times 10^{- 34} J s

4

2.11 \times 10^{- 34} J s

Explanation:

Energy absorbed during electron transition from
$n = 1$ to $n = 2$
$Δ E = 13.6 \times (\frac{1}{1^{2}} - \frac{1}{2^{2}}) e V = 10.2 e V$
As the hydrogen atom in the given problem also absorbs 10.2 $e V$ , the only possible electron transition is from $n = 1$ to $n = 2$ .
$∴$ Change in angular momentum
$Δ L = \frac{(2 - 1) h}{2 π} = \frac{6.6 \times 10^{- 34}}{2 \times 3.14} = 1.05 \times 10^{- 34} J s$

KCET - 2019