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PHXII12:ATOMS

356447 For the ground state the electron in the $H$ -atom has an angular momentum $h$ , according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,

1 Because angular momentum must be in the direction of spin of electron

2 Because Bohr model gives incorrect values of angular momentum

3 Because only one of these would have a minimum energy

4 Because electrons go around only in circular orbits

NCERT Exemplar

PHXII12:ATOMS

356448 The main defect of Bohr’s atomic model is

1 Can’t explain Zeeman effect

2 Can’t explain spectra of more complex atoms

3 Can’t explain fine structure of spectral lines

4 All of the above

PHXII12:ATOMS

356449 The energy of an electron in excited hydrogen atom is $- 3.4 e V$ . Then, according to Bohr's theory, the angular momentum of the electron is

1

2.1 \times 10^{- 34} J - s

2

3 \times 10^{- 34} J - s

3

2 \times 10^{- 34} J - s

4

0.5 \times 10^{- 34} J - s

Explanation:

According to the question,
energy, $E = - 3.4 e V$
This corresponds to $n = 2$ .
So, the angular momentum of the electron,
$m v r = \frac{n h}{2 π} = \frac{2 \times 6.626 \times 10^{- 34}}{2 \times 3.14}$
$= \frac{6.626 \times 10^{- 34}}{3.14}$
$= 2.1 \times 10^{- 34} J - s$

PHXII12:ATOMS

356450 For atomic model of hydrogen atom given by Niels Bohr, match the following proportionalities.
Column I
Column II
A
Angular momentum
P
$1 / n$
B
Velocity of electron
Q
$n^{2}$
C
Radius of electron
R
$1 / n^{2}$
D
Energy of electron
S
$n$

1 A-P, B-Q, C-R, D-S

2 A-S, B-R, C-Q, D-P

3 A-S, B-P, C-R, D-Q

4 A-S, B-P, C-Q, D-R

Explanation:

As we know, angular momentum
$(L) = \frac{n h}{2 π} \Rightarrow L \propto n$
Radius of an electron
$r_{n} = \frac{ε_{0} n^{2} h^{2}}{π m e^{2} z} \Rightarrow r_{n} \propto n^{2}$
Velocity of electrons
$v_{n} = \frac{z e^{2}}{2 ε_{0} n h} \Rightarrow v_{n} \propto 1 / n$
Potential energy i.e., $E_{n} = - \frac{m e^{4}}{4 n^{2} ε_{0}^{2} h^{2}}$
$E_{n} \propto 1 / n^{2}$
So option (4) is correct.

PHXII12:ATOMS

356447 For the ground state the electron in the $H$ -atom has an angular momentum $h$ , according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,

1 Because angular momentum must be in the direction of spin of electron

2 Because Bohr model gives incorrect values of angular momentum

3 Because only one of these would have a minimum energy

4 Because electrons go around only in circular orbits

NCERT Exemplar

PHXII12:ATOMS

356448 The main defect of Bohr’s atomic model is

1 Can’t explain Zeeman effect

2 Can’t explain spectra of more complex atoms

3 Can’t explain fine structure of spectral lines

4 All of the above

PHXII12:ATOMS

356449 The energy of an electron in excited hydrogen atom is $- 3.4 e V$ . Then, according to Bohr's theory, the angular momentum of the electron is

1

2.1 \times 10^{- 34} J - s

2

3 \times 10^{- 34} J - s

3

2 \times 10^{- 34} J - s

4

0.5 \times 10^{- 34} J - s

Explanation:

According to the question,
energy, $E = - 3.4 e V$
This corresponds to $n = 2$ .
So, the angular momentum of the electron,
$m v r = \frac{n h}{2 π} = \frac{2 \times 6.626 \times 10^{- 34}}{2 \times 3.14}$
$= \frac{6.626 \times 10^{- 34}}{3.14}$
$= 2.1 \times 10^{- 34} J - s$

PHXII12:ATOMS

356450 For atomic model of hydrogen atom given by Niels Bohr, match the following proportionalities.
Column I
Column II
A
Angular momentum
P
$1 / n$
B
Velocity of electron
Q
$n^{2}$
C
Radius of electron
R
$1 / n^{2}$
D
Energy of electron
S
$n$

1 A-P, B-Q, C-R, D-S

2 A-S, B-R, C-Q, D-P

3 A-S, B-P, C-R, D-Q

4 A-S, B-P, C-Q, D-R

Explanation:

As we know, angular momentum
$(L) = \frac{n h}{2 π} \Rightarrow L \propto n$
Radius of an electron
$r_{n} = \frac{ε_{0} n^{2} h^{2}}{π m e^{2} z} \Rightarrow r_{n} \propto n^{2}$
Velocity of electrons
$v_{n} = \frac{z e^{2}}{2 ε_{0} n h} \Rightarrow v_{n} \propto 1 / n$
Potential energy i.e., $E_{n} = - \frac{m e^{4}}{4 n^{2} ε_{0}^{2} h^{2}}$
$E_{n} \propto 1 / n^{2}$
So option (4) is correct.

PHXII12:ATOMS

356447 For the ground state the electron in the $H$ -atom has an angular momentum $h$ , according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,

1 Because angular momentum must be in the direction of spin of electron

2 Because Bohr model gives incorrect values of angular momentum

3 Because only one of these would have a minimum energy

4 Because electrons go around only in circular orbits

NCERT Exemplar

PHXII12:ATOMS

356448 The main defect of Bohr’s atomic model is

1 Can’t explain Zeeman effect

2 Can’t explain spectra of more complex atoms

3 Can’t explain fine structure of spectral lines

4 All of the above

PHXII12:ATOMS

356449 The energy of an electron in excited hydrogen atom is $- 3.4 e V$ . Then, according to Bohr's theory, the angular momentum of the electron is

1

2.1 \times 10^{- 34} J - s

2

3 \times 10^{- 34} J - s

3

2 \times 10^{- 34} J - s

4

0.5 \times 10^{- 34} J - s

Explanation:

According to the question,
energy, $E = - 3.4 e V$
This corresponds to $n = 2$ .
So, the angular momentum of the electron,
$m v r = \frac{n h}{2 π} = \frac{2 \times 6.626 \times 10^{- 34}}{2 \times 3.14}$
$= \frac{6.626 \times 10^{- 34}}{3.14}$
$= 2.1 \times 10^{- 34} J - s$

PHXII12:ATOMS

356450 For atomic model of hydrogen atom given by Niels Bohr, match the following proportionalities.
Column I
Column II
A
Angular momentum
P
$1 / n$
B
Velocity of electron
Q
$n^{2}$
C
Radius of electron
R
$1 / n^{2}$
D
Energy of electron
S
$n$

1 A-P, B-Q, C-R, D-S

2 A-S, B-R, C-Q, D-P

3 A-S, B-P, C-R, D-Q

4 A-S, B-P, C-Q, D-R

Explanation:

As we know, angular momentum
$(L) = \frac{n h}{2 π} \Rightarrow L \propto n$
Radius of an electron
$r_{n} = \frac{ε_{0} n^{2} h^{2}}{π m e^{2} z} \Rightarrow r_{n} \propto n^{2}$
Velocity of electrons
$v_{n} = \frac{z e^{2}}{2 ε_{0} n h} \Rightarrow v_{n} \propto 1 / n$
Potential energy i.e., $E_{n} = - \frac{m e^{4}}{4 n^{2} ε_{0}^{2} h^{2}}$
$E_{n} \propto 1 / n^{2}$
So option (4) is correct.

PHXII12:ATOMS

356447 For the ground state the electron in the $H$ -atom has an angular momentum $h$ , according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,

1 Because angular momentum must be in the direction of spin of electron

2 Because Bohr model gives incorrect values of angular momentum

3 Because only one of these would have a minimum energy

4 Because electrons go around only in circular orbits

NCERT Exemplar

PHXII12:ATOMS

356448 The main defect of Bohr’s atomic model is

1 Can’t explain Zeeman effect

2 Can’t explain spectra of more complex atoms

3 Can’t explain fine structure of spectral lines

4 All of the above

PHXII12:ATOMS

356449 The energy of an electron in excited hydrogen atom is $- 3.4 e V$ . Then, according to Bohr's theory, the angular momentum of the electron is

1

2.1 \times 10^{- 34} J - s

2

3 \times 10^{- 34} J - s

3

2 \times 10^{- 34} J - s

4

0.5 \times 10^{- 34} J - s

Explanation:

According to the question,
energy, $E = - 3.4 e V$
This corresponds to $n = 2$ .
So, the angular momentum of the electron,
$m v r = \frac{n h}{2 π} = \frac{2 \times 6.626 \times 10^{- 34}}{2 \times 3.14}$
$= \frac{6.626 \times 10^{- 34}}{3.14}$
$= 2.1 \times 10^{- 34} J - s$

PHXII12:ATOMS

356450 For atomic model of hydrogen atom given by Niels Bohr, match the following proportionalities.
Column I
Column II
A
Angular momentum
P
$1 / n$
B
Velocity of electron
Q
$n^{2}$
C
Radius of electron
R
$1 / n^{2}$
D
Energy of electron
S
$n$

1 A-P, B-Q, C-R, D-S

2 A-S, B-R, C-Q, D-P

3 A-S, B-P, C-R, D-Q

4 A-S, B-P, C-Q, D-R

Explanation:

As we know, angular momentum
$(L) = \frac{n h}{2 π} \Rightarrow L \propto n$
Radius of an electron
$r_{n} = \frac{ε_{0} n^{2} h^{2}}{π m e^{2} z} \Rightarrow r_{n} \propto n^{2}$
Velocity of electrons
$v_{n} = \frac{z e^{2}}{2 ε_{0} n h} \Rightarrow v_{n} \propto 1 / n$
Potential energy i.e., $E_{n} = - \frac{m e^{4}}{4 n^{2} ε_{0}^{2} h^{2}}$
$E_{n} \propto 1 / n^{2}$
So option (4) is correct.