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ATOMS

145307 The radius of inner most orbit of hydrogen atom is $5.3 \times 10^{- 11} m$ . What is the radius of third allowed orbit of hydrogen atom?

1

4.77 \overset{\circ}{\AA}

2

0.53 \AA

3

1.06 \overset{\circ}{\AA}

4

1.59 \AA

Explanation:

A The radius of inner most orbit of $H$ -atom is -
$r_{1} = 5.3 \times 10^{- 11} m$
$r_{1} = \frac{0.529}{z}$
$r_{3} = \frac{0.529}{z} (3)^{2}$
$r_{3} = r_{1} \times 9$
$= 9 \times 0.53$
$r_{3} = 4.77 \overset{\circ}{A}$

NEET (UG) 07.05.2023

ATOMS

145308 The angular momentum for the electron in Bohr's orbit is $L$ . If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be:

1

\frac{L}{2}

2 zero

3

L

4

2 L

Explanation:

C We know that, by conservation of angular momentum, $L =$ mvr,
In Bohr's model the radius and velocity of orbit-
$r \propto n^{2}, v \propto \frac{1}{n}$
$∴ E \propto n$
By Bohr's postulate-
Also, $L = \frac{nh}{2 π}$
Bohr's orbit is, $L_{1} = L = \frac{1 . h}{2 π}$
$L_{2} = 2 [L] = 2 L$
$L_{2} = \frac{2 h}{2 π}$
Then change in total angular momentum $= L_{2} - L_{1} = 2 L$ $- L = L$

JEE Main-10.04.2023

ATOMS

145309 A small particle of mass $m$ moves in such a way that its potential energy $U = \frac{1}{2} m ω^{2} r^{2}$ where $ω$ is constant and $r$ is the distance of the particle from origin Assuming Bohr's quantization of momentum and circular orbit, the radius of $n^{th}$ orbit will be proportional to

1

\sqrt{n}

2

n

3

n^{2}

4

\frac{1}{n}

Explanation:

A Given potential energy $U = \frac{1}{2} m ω^{2} r^{2}$
The relation between force and potential energy
$F = - \frac{dU}{dr} = - m ω^{2} r$
This force equal to centrifugal force, $f = \frac{m v^{2}}{r}$
Now $m ω^{2} r = \frac{{mv}^{2}}{r} \Rightarrow v = ω r$
From conservation of angular momentum-
$mvr = \frac{nh}{2 π}$
From (i) \(ii)
${m ω r}^{2} = \frac{n h}{2 π}$
$r^{2} \propto n$
$\Rightarrow r \propto \sqrt{n}$

JEE Main-06.04.2023

ATOMS

145310 The radius of electron's second stationary orbit in Bohr's atom is $R$ . The radius of 3rd orbit will be:

1

\frac{R}{3}

2

9 R

3

2.25 R

4

3 R

Explanation:

C According to Bohr's radius orbit
$R_{n} \propto \frac{n^{2}}{Z}$
So, $R_{n} \propto n^{2}$
$R_{2} \propto (2)^{2}$
$R_{3} \propto (3)^{2}$
Dividing equation (i) by equation (ii), we get -
$\frac{R_{3}}{R_{2}} = \frac{9}{4}$
$R_{3} = 2.25 \times R$
$R_{3} = 2.25 R$

JEE Main-31.01.2023

ATOMS

145311 Speed of an electron 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

3.6 \times 10^{6}

2

7.5 \times 10^{6}

3

1.8 \times 10^{6}

4

8.4 \times 10^{6}

Explanation:

D Given that, $v_{7} = 3.6 \times 10^{6} m / s$
We know that,
$v_{n} \propto \frac{Z}{n}$
For $Z = 1, v_{n} \propto \frac{1}{n}$ $∴ \frac{v_{3}}{v_{7}} = \frac{7}{3}$
$v_{3} = 3.6 \times 10^{6} \times \frac{7}{3}$
$v_{3} = 8.4 \times 10^{6} m / \sec$

JEE Main-30.01.2023

ATOMS

145307 The radius of inner most orbit of hydrogen atom is $5.3 \times 10^{- 11} m$ . What is the radius of third allowed orbit of hydrogen atom?

1

4.77 \overset{\circ}{\AA}

2

0.53 \AA

3

1.06 \overset{\circ}{\AA}

4

1.59 \AA

Explanation:

A The radius of inner most orbit of $H$ -atom is -
$r_{1} = 5.3 \times 10^{- 11} m$
$r_{1} = \frac{0.529}{z}$
$r_{3} = \frac{0.529}{z} (3)^{2}$
$r_{3} = r_{1} \times 9$
$= 9 \times 0.53$
$r_{3} = 4.77 \overset{\circ}{A}$

NEET (UG) 07.05.2023

ATOMS

145308 The angular momentum for the electron in Bohr's orbit is $L$ . If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be:

1

\frac{L}{2}

2 zero

3

L

4

2 L

Explanation:

C We know that, by conservation of angular momentum, $L =$ mvr,
In Bohr's model the radius and velocity of orbit-
$r \propto n^{2}, v \propto \frac{1}{n}$
$∴ E \propto n$
By Bohr's postulate-
Also, $L = \frac{nh}{2 π}$
Bohr's orbit is, $L_{1} = L = \frac{1 . h}{2 π}$
$L_{2} = 2 [L] = 2 L$
$L_{2} = \frac{2 h}{2 π}$
Then change in total angular momentum $= L_{2} - L_{1} = 2 L$ $- L = L$

JEE Main-10.04.2023

ATOMS

145309 A small particle of mass $m$ moves in such a way that its potential energy $U = \frac{1}{2} m ω^{2} r^{2}$ where $ω$ is constant and $r$ is the distance of the particle from origin Assuming Bohr's quantization of momentum and circular orbit, the radius of $n^{th}$ orbit will be proportional to

1

\sqrt{n}

2

n

3

n^{2}

4

\frac{1}{n}

Explanation:

A Given potential energy $U = \frac{1}{2} m ω^{2} r^{2}$
The relation between force and potential energy
$F = - \frac{dU}{dr} = - m ω^{2} r$
This force equal to centrifugal force, $f = \frac{m v^{2}}{r}$
Now $m ω^{2} r = \frac{{mv}^{2}}{r} \Rightarrow v = ω r$
From conservation of angular momentum-
$mvr = \frac{nh}{2 π}$
From (i) \(ii)
${m ω r}^{2} = \frac{n h}{2 π}$
$r^{2} \propto n$
$\Rightarrow r \propto \sqrt{n}$

JEE Main-06.04.2023

ATOMS

145310 The radius of electron's second stationary orbit in Bohr's atom is $R$ . The radius of 3rd orbit will be:

1

\frac{R}{3}

2

9 R

3

2.25 R

4

3 R

Explanation:

C According to Bohr's radius orbit
$R_{n} \propto \frac{n^{2}}{Z}$
So, $R_{n} \propto n^{2}$
$R_{2} \propto (2)^{2}$
$R_{3} \propto (3)^{2}$
Dividing equation (i) by equation (ii), we get -
$\frac{R_{3}}{R_{2}} = \frac{9}{4}$
$R_{3} = 2.25 \times R$
$R_{3} = 2.25 R$

JEE Main-31.01.2023

ATOMS

145311 Speed of an electron 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

3.6 \times 10^{6}

2

7.5 \times 10^{6}

3

1.8 \times 10^{6}

4

8.4 \times 10^{6}

Explanation:

D Given that, $v_{7} = 3.6 \times 10^{6} m / s$
We know that,
$v_{n} \propto \frac{Z}{n}$
For $Z = 1, v_{n} \propto \frac{1}{n}$ $∴ \frac{v_{3}}{v_{7}} = \frac{7}{3}$
$v_{3} = 3.6 \times 10^{6} \times \frac{7}{3}$
$v_{3} = 8.4 \times 10^{6} m / \sec$

JEE Main-30.01.2023

ATOMS

145307 The radius of inner most orbit of hydrogen atom is $5.3 \times 10^{- 11} m$ . What is the radius of third allowed orbit of hydrogen atom?

1

4.77 \overset{\circ}{\AA}

2

0.53 \AA

3

1.06 \overset{\circ}{\AA}

4

1.59 \AA

Explanation:

A The radius of inner most orbit of $H$ -atom is -
$r_{1} = 5.3 \times 10^{- 11} m$
$r_{1} = \frac{0.529}{z}$
$r_{3} = \frac{0.529}{z} (3)^{2}$
$r_{3} = r_{1} \times 9$
$= 9 \times 0.53$
$r_{3} = 4.77 \overset{\circ}{A}$

NEET (UG) 07.05.2023

ATOMS

145308 The angular momentum for the electron in Bohr's orbit is $L$ . If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be:

1

\frac{L}{2}

2 zero

3

L

4

2 L

Explanation:

C We know that, by conservation of angular momentum, $L =$ mvr,
In Bohr's model the radius and velocity of orbit-
$r \propto n^{2}, v \propto \frac{1}{n}$
$∴ E \propto n$
By Bohr's postulate-
Also, $L = \frac{nh}{2 π}$
Bohr's orbit is, $L_{1} = L = \frac{1 . h}{2 π}$
$L_{2} = 2 [L] = 2 L$
$L_{2} = \frac{2 h}{2 π}$
Then change in total angular momentum $= L_{2} - L_{1} = 2 L$ $- L = L$

JEE Main-10.04.2023

ATOMS

145309 A small particle of mass $m$ moves in such a way that its potential energy $U = \frac{1}{2} m ω^{2} r^{2}$ where $ω$ is constant and $r$ is the distance of the particle from origin Assuming Bohr's quantization of momentum and circular orbit, the radius of $n^{th}$ orbit will be proportional to

1

\sqrt{n}

2

n

3

n^{2}

4

\frac{1}{n}

Explanation:

A Given potential energy $U = \frac{1}{2} m ω^{2} r^{2}$
The relation between force and potential energy
$F = - \frac{dU}{dr} = - m ω^{2} r$
This force equal to centrifugal force, $f = \frac{m v^{2}}{r}$
Now $m ω^{2} r = \frac{{mv}^{2}}{r} \Rightarrow v = ω r$
From conservation of angular momentum-
$mvr = \frac{nh}{2 π}$
From (i) \(ii)
${m ω r}^{2} = \frac{n h}{2 π}$
$r^{2} \propto n$
$\Rightarrow r \propto \sqrt{n}$

JEE Main-06.04.2023

ATOMS

145310 The radius of electron's second stationary orbit in Bohr's atom is $R$ . The radius of 3rd orbit will be:

1

\frac{R}{3}

2

9 R

3

2.25 R

4

3 R

Explanation:

C According to Bohr's radius orbit
$R_{n} \propto \frac{n^{2}}{Z}$
So, $R_{n} \propto n^{2}$
$R_{2} \propto (2)^{2}$
$R_{3} \propto (3)^{2}$
Dividing equation (i) by equation (ii), we get -
$\frac{R_{3}}{R_{2}} = \frac{9}{4}$
$R_{3} = 2.25 \times R$
$R_{3} = 2.25 R$

JEE Main-31.01.2023

ATOMS

145311 Speed of an electron 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

3.6 \times 10^{6}

2

7.5 \times 10^{6}

3

1.8 \times 10^{6}

4

8.4 \times 10^{6}

Explanation:

D Given that, $v_{7} = 3.6 \times 10^{6} m / s$
We know that,
$v_{n} \propto \frac{Z}{n}$
For $Z = 1, v_{n} \propto \frac{1}{n}$ $∴ \frac{v_{3}}{v_{7}} = \frac{7}{3}$
$v_{3} = 3.6 \times 10^{6} \times \frac{7}{3}$
$v_{3} = 8.4 \times 10^{6} m / \sec$

JEE Main-30.01.2023

ATOMS

145307 The radius of inner most orbit of hydrogen atom is $5.3 \times 10^{- 11} m$ . What is the radius of third allowed orbit of hydrogen atom?

1

4.77 \overset{\circ}{\AA}

2

0.53 \AA

3

1.06 \overset{\circ}{\AA}

4

1.59 \AA

Explanation:

A The radius of inner most orbit of $H$ -atom is -
$r_{1} = 5.3 \times 10^{- 11} m$
$r_{1} = \frac{0.529}{z}$
$r_{3} = \frac{0.529}{z} (3)^{2}$
$r_{3} = r_{1} \times 9$
$= 9 \times 0.53$
$r_{3} = 4.77 \overset{\circ}{A}$

NEET (UG) 07.05.2023

ATOMS

145308 The angular momentum for the electron in Bohr's orbit is $L$ . If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be:

1

\frac{L}{2}

2 zero

3

L

4

2 L

Explanation:

C We know that, by conservation of angular momentum, $L =$ mvr,
In Bohr's model the radius and velocity of orbit-
$r \propto n^{2}, v \propto \frac{1}{n}$
$∴ E \propto n$
By Bohr's postulate-
Also, $L = \frac{nh}{2 π}$
Bohr's orbit is, $L_{1} = L = \frac{1 . h}{2 π}$
$L_{2} = 2 [L] = 2 L$
$L_{2} = \frac{2 h}{2 π}$
Then change in total angular momentum $= L_{2} - L_{1} = 2 L$ $- L = L$

JEE Main-10.04.2023

ATOMS

145309 A small particle of mass $m$ moves in such a way that its potential energy $U = \frac{1}{2} m ω^{2} r^{2}$ where $ω$ is constant and $r$ is the distance of the particle from origin Assuming Bohr's quantization of momentum and circular orbit, the radius of $n^{th}$ orbit will be proportional to

1

\sqrt{n}

2

n

3

n^{2}

4

\frac{1}{n}

Explanation:

A Given potential energy $U = \frac{1}{2} m ω^{2} r^{2}$
The relation between force and potential energy
$F = - \frac{dU}{dr} = - m ω^{2} r$
This force equal to centrifugal force, $f = \frac{m v^{2}}{r}$
Now $m ω^{2} r = \frac{{mv}^{2}}{r} \Rightarrow v = ω r$
From conservation of angular momentum-
$mvr = \frac{nh}{2 π}$
From (i) \(ii)
${m ω r}^{2} = \frac{n h}{2 π}$
$r^{2} \propto n$
$\Rightarrow r \propto \sqrt{n}$

JEE Main-06.04.2023

ATOMS

145310 The radius of electron's second stationary orbit in Bohr's atom is $R$ . The radius of 3rd orbit will be:

1

\frac{R}{3}

2

9 R

3

2.25 R

4

3 R

Explanation:

C According to Bohr's radius orbit
$R_{n} \propto \frac{n^{2}}{Z}$
So, $R_{n} \propto n^{2}$
$R_{2} \propto (2)^{2}$
$R_{3} \propto (3)^{2}$
Dividing equation (i) by equation (ii), we get -
$\frac{R_{3}}{R_{2}} = \frac{9}{4}$
$R_{3} = 2.25 \times R$
$R_{3} = 2.25 R$

JEE Main-31.01.2023

ATOMS

145311 Speed of an electron 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

3.6 \times 10^{6}

2

7.5 \times 10^{6}

3

1.8 \times 10^{6}

4

8.4 \times 10^{6}

Explanation:

D Given that, $v_{7} = 3.6 \times 10^{6} m / s$
We know that,
$v_{n} \propto \frac{Z}{n}$
For $Z = 1, v_{n} \propto \frac{1}{n}$ $∴ \frac{v_{3}}{v_{7}} = \frac{7}{3}$
$v_{3} = 3.6 \times 10^{6} \times \frac{7}{3}$
$v_{3} = 8.4 \times 10^{6} m / \sec$

JEE Main-30.01.2023

ATOMS

145307 The radius of inner most orbit of hydrogen atom is $5.3 \times 10^{- 11} m$ . What is the radius of third allowed orbit of hydrogen atom?

1

4.77 \overset{\circ}{\AA}

2

0.53 \AA

3

1.06 \overset{\circ}{\AA}

4

1.59 \AA

Explanation:

A The radius of inner most orbit of $H$ -atom is -
$r_{1} = 5.3 \times 10^{- 11} m$
$r_{1} = \frac{0.529}{z}$
$r_{3} = \frac{0.529}{z} (3)^{2}$
$r_{3} = r_{1} \times 9$
$= 9 \times 0.53$
$r_{3} = 4.77 \overset{\circ}{A}$

NEET (UG) 07.05.2023

ATOMS

145308 The angular momentum for the electron in Bohr's orbit is $L$ . If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be:

1

\frac{L}{2}

2 zero

3

L

4

2 L

Explanation:

C We know that, by conservation of angular momentum, $L =$ mvr,
In Bohr's model the radius and velocity of orbit-
$r \propto n^{2}, v \propto \frac{1}{n}$
$∴ E \propto n$
By Bohr's postulate-
Also, $L = \frac{nh}{2 π}$
Bohr's orbit is, $L_{1} = L = \frac{1 . h}{2 π}$
$L_{2} = 2 [L] = 2 L$
$L_{2} = \frac{2 h}{2 π}$
Then change in total angular momentum $= L_{2} - L_{1} = 2 L$ $- L = L$

JEE Main-10.04.2023

ATOMS

145309 A small particle of mass $m$ moves in such a way that its potential energy $U = \frac{1}{2} m ω^{2} r^{2}$ where $ω$ is constant and $r$ is the distance of the particle from origin Assuming Bohr's quantization of momentum and circular orbit, the radius of $n^{th}$ orbit will be proportional to

1

\sqrt{n}

2

n

3

n^{2}

4

\frac{1}{n}

Explanation:

A Given potential energy $U = \frac{1}{2} m ω^{2} r^{2}$
The relation between force and potential energy
$F = - \frac{dU}{dr} = - m ω^{2} r$
This force equal to centrifugal force, $f = \frac{m v^{2}}{r}$
Now $m ω^{2} r = \frac{{mv}^{2}}{r} \Rightarrow v = ω r$
From conservation of angular momentum-
$mvr = \frac{nh}{2 π}$
From (i) \(ii)
${m ω r}^{2} = \frac{n h}{2 π}$
$r^{2} \propto n$
$\Rightarrow r \propto \sqrt{n}$

JEE Main-06.04.2023

ATOMS

145310 The radius of electron's second stationary orbit in Bohr's atom is $R$ . The radius of 3rd orbit will be:

1

\frac{R}{3}

2

9 R

3

2.25 R

4

3 R

Explanation:

C According to Bohr's radius orbit
$R_{n} \propto \frac{n^{2}}{Z}$
So, $R_{n} \propto n^{2}$
$R_{2} \propto (2)^{2}$
$R_{3} \propto (3)^{2}$
Dividing equation (i) by equation (ii), we get -
$\frac{R_{3}}{R_{2}} = \frac{9}{4}$
$R_{3} = 2.25 \times R$
$R_{3} = 2.25 R$

JEE Main-31.01.2023

ATOMS

145311 Speed of an electron 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

3.6 \times 10^{6}

2

7.5 \times 10^{6}

3

1.8 \times 10^{6}

4

8.4 \times 10^{6}

Explanation:

D Given that, $v_{7} = 3.6 \times 10^{6} m / s$
We know that,
$v_{n} \propto \frac{Z}{n}$
For $Z = 1, v_{n} \propto \frac{1}{n}$ $∴ \frac{v_{3}}{v_{7}} = \frac{7}{3}$
$v_{3} = 3.6 \times 10^{6} \times \frac{7}{3}$
$v_{3} = 8.4 \times 10^{6} m / \sec$

JEE Main-30.01.2023

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