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PHXII12:ATOMS

356404 For electron moving in $n^{t h}$ orbit of the atom, the angular velocity is proportional to:

1

n

2

\frac{1}{n}

3

n^{3}

4

\frac{1}{n^{3}}

Explanation:

$v α \frac{1}{n} & r α n^{2}$
$v = r ω \Rightarrow ω = \frac{v}{r}$
$ω α \frac{1}{n^{3}}$

PHXII12:ATOMS

356405 An electron is moving in an orbit of a hydrogen atom from which there can be a maximum of six transitions. An electron is moving in an orbit of another hydrogen atom from which there can be a maximum of three transitions. The ratio of the velocity of the electron in these two orbits is

1

1 / 2

2

2 / 1

3

5 / 4

4

3 / 4

Explanation:

Number of spectral lines obtained due to transition of electron from $n^{t h}$ orbit to lower orbits is
$N = \frac{n (n - 1)}{2}$
In first case, $N = 6$
$∴ 6 = n \frac{(n - 1)}{2} \Rightarrow n = 4$
In the second case, $N = 3$
$∴ 3 = n \frac{(n - 1)}{2} \Rightarrow n = 3$
Velocity of electron in hydrogen atom in $n^{t h}$ orbit is
$v_{n} = \frac{e^{2}}{2 ε_{0} n h} \Rightarrow v_{n} \propto \frac{1}{n}$
$∴ \frac{v_{4}}{v_{3}} = \frac{3}{4}$

KCET - 2010

PHXII12:ATOMS

356406 A photon of energy 12.09 $e V$ is absorbed by an electron in ground state of a hydrogen atom. What will be the energy level of electron? The energy of electron in the ground state of hydrogen atom is $-$ 13.6 $e V$ .

1 3

2 6

3 8

4 4

Explanation:

$E_{n} - E_{1} = h f$
or $E_{n} = h f + E_{1} = 12.09 - 13.6 = - 1.51 e V$
As $E_{n} = \frac{E_{1}}{n^{2}} ∴ n^{2} = \frac{E_{1}}{E_{n}} = \frac{- 13.6}{- 1.51} ≃ 9$
or $n = 3$

PHXII12:ATOMS

356407 In which of the following systems will the radius of the first orbit $(n = 1$ ) be minimum?

1 Deuterium atom

2 Hydrogen atom

3 Doubly ionised lithium

4 Singly ionised helium

Explanation:

Radius of $n$ th Bohr's orbit of
hydrogen like atom,
$r_{n} \propto \frac{n^{2}}{Z}$
where, $Z$ is the atomic number.
For first orbit.
$n = 1$
$∴ r_{n} \propto \frac{1}{Z}$
Now, for radius of 1 st orbit to be minimum, its atomic number must be greater. Here, in the problem, atomic number $Z$ is greater for double ionised lithium, $i . e . Z = 3$ . Hence, for the radius of 1 st orbit will be minimum.

PHXII12:ATOMS

356404 For electron moving in $n^{t h}$ orbit of the atom, the angular velocity is proportional to:

1

n

2

\frac{1}{n}

3

n^{3}

4

\frac{1}{n^{3}}

Explanation:

$v α \frac{1}{n} & r α n^{2}$
$v = r ω \Rightarrow ω = \frac{v}{r}$
$ω α \frac{1}{n^{3}}$

PHXII12:ATOMS

356405 An electron is moving in an orbit of a hydrogen atom from which there can be a maximum of six transitions. An electron is moving in an orbit of another hydrogen atom from which there can be a maximum of three transitions. The ratio of the velocity of the electron in these two orbits is

1

1 / 2

2

2 / 1

3

5 / 4

4

3 / 4

Explanation:

Number of spectral lines obtained due to transition of electron from $n^{t h}$ orbit to lower orbits is
$N = \frac{n (n - 1)}{2}$
In first case, $N = 6$
$∴ 6 = n \frac{(n - 1)}{2} \Rightarrow n = 4$
In the second case, $N = 3$
$∴ 3 = n \frac{(n - 1)}{2} \Rightarrow n = 3$
Velocity of electron in hydrogen atom in $n^{t h}$ orbit is
$v_{n} = \frac{e^{2}}{2 ε_{0} n h} \Rightarrow v_{n} \propto \frac{1}{n}$
$∴ \frac{v_{4}}{v_{3}} = \frac{3}{4}$

KCET - 2010

PHXII12:ATOMS

356406 A photon of energy 12.09 $e V$ is absorbed by an electron in ground state of a hydrogen atom. What will be the energy level of electron? The energy of electron in the ground state of hydrogen atom is $-$ 13.6 $e V$ .

1 3

2 6

3 8

4 4

Explanation:

$E_{n} - E_{1} = h f$
or $E_{n} = h f + E_{1} = 12.09 - 13.6 = - 1.51 e V$
As $E_{n} = \frac{E_{1}}{n^{2}} ∴ n^{2} = \frac{E_{1}}{E_{n}} = \frac{- 13.6}{- 1.51} ≃ 9$
or $n = 3$

PHXII12:ATOMS

356407 In which of the following systems will the radius of the first orbit $(n = 1$ ) be minimum?

1 Deuterium atom

2 Hydrogen atom

3 Doubly ionised lithium

4 Singly ionised helium

Explanation:

Radius of $n$ th Bohr's orbit of
hydrogen like atom,
$r_{n} \propto \frac{n^{2}}{Z}$
where, $Z$ is the atomic number.
For first orbit.
$n = 1$
$∴ r_{n} \propto \frac{1}{Z}$
Now, for radius of 1 st orbit to be minimum, its atomic number must be greater. Here, in the problem, atomic number $Z$ is greater for double ionised lithium, $i . e . Z = 3$ . Hence, for the radius of 1 st orbit will be minimum.

PHXII12:ATOMS

356404 For electron moving in $n^{t h}$ orbit of the atom, the angular velocity is proportional to:

1

n

2

\frac{1}{n}

3

n^{3}

4

\frac{1}{n^{3}}

Explanation:

$v α \frac{1}{n} & r α n^{2}$
$v = r ω \Rightarrow ω = \frac{v}{r}$
$ω α \frac{1}{n^{3}}$

PHXII12:ATOMS

356405 An electron is moving in an orbit of a hydrogen atom from which there can be a maximum of six transitions. An electron is moving in an orbit of another hydrogen atom from which there can be a maximum of three transitions. The ratio of the velocity of the electron in these two orbits is

1

1 / 2

2

2 / 1

3

5 / 4

4

3 / 4

Explanation:

Number of spectral lines obtained due to transition of electron from $n^{t h}$ orbit to lower orbits is
$N = \frac{n (n - 1)}{2}$
In first case, $N = 6$
$∴ 6 = n \frac{(n - 1)}{2} \Rightarrow n = 4$
In the second case, $N = 3$
$∴ 3 = n \frac{(n - 1)}{2} \Rightarrow n = 3$
Velocity of electron in hydrogen atom in $n^{t h}$ orbit is
$v_{n} = \frac{e^{2}}{2 ε_{0} n h} \Rightarrow v_{n} \propto \frac{1}{n}$
$∴ \frac{v_{4}}{v_{3}} = \frac{3}{4}$

KCET - 2010

PHXII12:ATOMS

356406 A photon of energy 12.09 $e V$ is absorbed by an electron in ground state of a hydrogen atom. What will be the energy level of electron? The energy of electron in the ground state of hydrogen atom is $-$ 13.6 $e V$ .

1 3

2 6

3 8

4 4

Explanation:

$E_{n} - E_{1} = h f$
or $E_{n} = h f + E_{1} = 12.09 - 13.6 = - 1.51 e V$
As $E_{n} = \frac{E_{1}}{n^{2}} ∴ n^{2} = \frac{E_{1}}{E_{n}} = \frac{- 13.6}{- 1.51} ≃ 9$
or $n = 3$

PHXII12:ATOMS

356407 In which of the following systems will the radius of the first orbit $(n = 1$ ) be minimum?

1 Deuterium atom

2 Hydrogen atom

3 Doubly ionised lithium

4 Singly ionised helium

Explanation:

Radius of $n$ th Bohr's orbit of
hydrogen like atom,
$r_{n} \propto \frac{n^{2}}{Z}$
where, $Z$ is the atomic number.
For first orbit.
$n = 1$
$∴ r_{n} \propto \frac{1}{Z}$
Now, for radius of 1 st orbit to be minimum, its atomic number must be greater. Here, in the problem, atomic number $Z$ is greater for double ionised lithium, $i . e . Z = 3$ . Hence, for the radius of 1 st orbit will be minimum.

PHXII12:ATOMS

356404 For electron moving in $n^{t h}$ orbit of the atom, the angular velocity is proportional to:

1

n

2

\frac{1}{n}

3

n^{3}

4

\frac{1}{n^{3}}

Explanation:

$v α \frac{1}{n} & r α n^{2}$
$v = r ω \Rightarrow ω = \frac{v}{r}$
$ω α \frac{1}{n^{3}}$

PHXII12:ATOMS

356405 An electron is moving in an orbit of a hydrogen atom from which there can be a maximum of six transitions. An electron is moving in an orbit of another hydrogen atom from which there can be a maximum of three transitions. The ratio of the velocity of the electron in these two orbits is

1

1 / 2

2

2 / 1

3

5 / 4

4

3 / 4

Explanation:

Number of spectral lines obtained due to transition of electron from $n^{t h}$ orbit to lower orbits is
$N = \frac{n (n - 1)}{2}$
In first case, $N = 6$
$∴ 6 = n \frac{(n - 1)}{2} \Rightarrow n = 4$
In the second case, $N = 3$
$∴ 3 = n \frac{(n - 1)}{2} \Rightarrow n = 3$
Velocity of electron in hydrogen atom in $n^{t h}$ orbit is
$v_{n} = \frac{e^{2}}{2 ε_{0} n h} \Rightarrow v_{n} \propto \frac{1}{n}$
$∴ \frac{v_{4}}{v_{3}} = \frac{3}{4}$

KCET - 2010

PHXII12:ATOMS

356406 A photon of energy 12.09 $e V$ is absorbed by an electron in ground state of a hydrogen atom. What will be the energy level of electron? The energy of electron in the ground state of hydrogen atom is $-$ 13.6 $e V$ .

1 3

2 6

3 8

4 4

Explanation:

$E_{n} - E_{1} = h f$
or $E_{n} = h f + E_{1} = 12.09 - 13.6 = - 1.51 e V$
As $E_{n} = \frac{E_{1}}{n^{2}} ∴ n^{2} = \frac{E_{1}}{E_{n}} = \frac{- 13.6}{- 1.51} ≃ 9$
or $n = 3$

PHXII12:ATOMS

356407 In which of the following systems will the radius of the first orbit $(n = 1$ ) be minimum?

1 Deuterium atom

2 Hydrogen atom

3 Doubly ionised lithium

4 Singly ionised helium

Explanation:

Radius of $n$ th Bohr's orbit of
hydrogen like atom,
$r_{n} \propto \frac{n^{2}}{Z}$
where, $Z$ is the atomic number.
For first orbit.
$n = 1$
$∴ r_{n} \propto \frac{1}{Z}$
Now, for radius of 1 st orbit to be minimum, its atomic number must be greater. Here, in the problem, atomic number $Z$ is greater for double ionised lithium, $i . e . Z = 3$ . Hence, for the radius of 1 st orbit will be minimum.