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PHXII12:ATOMS

356508 As per Bohr model the minimum energy (in $e V$ ) required to remove the electron from the ground state of doubly ionised lithium $(Z = 3)$ is

1

1.51

2

13.6

3

40.8

4

122.4

Explanation:

Ionsiation energy
$= Z^{2} R h c (\frac{1}{{(1)}^{2}} - \frac{1}{{(\infty)}^{2}})$
$= Z^{2} R h c = {(3)}^{2} \times 13.6 e V = 122.4 e V$

PHXII12:ATOMS

356509 An electron of a hydrogen like atom, having $Z = 4$ , jumps from $4^{th}$ energy state to $2^{nd}$ energy state. The energy released in this process, will be (Given $R c h = 13.6 e V$ ) Where, $R =$ Rydberg constant, $c =$ Speed of light in vacuum, $h =$ Planck's constant

1

40.8 e V

2

13.6 e V

3

3.4 e V

4

10.5 e V

Explanation:

As, $Z = 4, n_{f} = 2, n_{i} = 4$
The energy released is given by
$Δ E = 13.6 Z^{2} [\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}] = 13.6 \times 4^{2} [\frac{1}{4} - \frac{1}{16}]$
$Δ E = 13.6 \times 16 [\frac{4 - 1}{16}]$
$= 13.6 \times 3 = 40.8 e V$

JEE - 2023

PHXII12:ATOMS

356510 For which one of the following, Bohr model is not valid?

1 Singly ionized helium atom

(H e^{+})

2 Deuteron atom

3 Singly ionized neon atom

(N e^{+})

4 Hydrogen atom

NEET - 2020

PHXII12:ATOMS

356511 Assertion :
There is a difference in the spectrum of sunlight when passed through "yellow-glass" and that of the light from a sodium vapour lamp.
Reason :
The sunlight always gives continuous spectrum with absorption lines (the Fraunhofer lines).

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII12:ATOMS

356512 Which energy state of the triply ionised beryllium has the same electron orbital radius as that of ground state of hydrogen? Given $Z$ for $B e = 4.$

1

n = 1

2

n = 2

3

n = 3

4

n = 4

Explanation:

For an atom of atomic number $Z$ , radius of $n^{t h}$ orbit is given by
$r_{n} = \frac{k n^{2}}{Z} (1)$
where $k = c o n s t a n t$
For ground state of hydrogen, $Z = 1, n = 1,$ so that $r_{1} = k$
Let $n$ be the energy state of $B e^{+ 3}$ for which orbital radius is $r_{1} .$
Put $Z = 4$ and $r_{n} = r_{1} = k$ in Eq.(1)
$r_{1} = \frac{r_{1} n^{2}}{4} \Rightarrow n = 2$

PHXII12:ATOMS

356508 As per Bohr model the minimum energy (in $e V$ ) required to remove the electron from the ground state of doubly ionised lithium $(Z = 3)$ is

1

1.51

2

13.6

3

40.8

4

122.4

Explanation:

Ionsiation energy
$= Z^{2} R h c (\frac{1}{{(1)}^{2}} - \frac{1}{{(\infty)}^{2}})$
$= Z^{2} R h c = {(3)}^{2} \times 13.6 e V = 122.4 e V$

PHXII12:ATOMS

356509 An electron of a hydrogen like atom, having $Z = 4$ , jumps from $4^{th}$ energy state to $2^{nd}$ energy state. The energy released in this process, will be (Given $R c h = 13.6 e V$ ) Where, $R =$ Rydberg constant, $c =$ Speed of light in vacuum, $h =$ Planck's constant

1

40.8 e V

2

13.6 e V

3

3.4 e V

4

10.5 e V

Explanation:

As, $Z = 4, n_{f} = 2, n_{i} = 4$
The energy released is given by
$Δ E = 13.6 Z^{2} [\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}] = 13.6 \times 4^{2} [\frac{1}{4} - \frac{1}{16}]$
$Δ E = 13.6 \times 16 [\frac{4 - 1}{16}]$
$= 13.6 \times 3 = 40.8 e V$

JEE - 2023

PHXII12:ATOMS

356510 For which one of the following, Bohr model is not valid?

1 Singly ionized helium atom

(H e^{+})

2 Deuteron atom

3 Singly ionized neon atom

(N e^{+})

4 Hydrogen atom

NEET - 2020

PHXII12:ATOMS

356511 Assertion :
There is a difference in the spectrum of sunlight when passed through "yellow-glass" and that of the light from a sodium vapour lamp.
Reason :
The sunlight always gives continuous spectrum with absorption lines (the Fraunhofer lines).

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII12:ATOMS

356512 Which energy state of the triply ionised beryllium has the same electron orbital radius as that of ground state of hydrogen? Given $Z$ for $B e = 4.$

1

n = 1

2

n = 2

3

n = 3

4

n = 4

Explanation:

For an atom of atomic number $Z$ , radius of $n^{t h}$ orbit is given by
$r_{n} = \frac{k n^{2}}{Z} (1)$
where $k = c o n s t a n t$
For ground state of hydrogen, $Z = 1, n = 1,$ so that $r_{1} = k$
Let $n$ be the energy state of $B e^{+ 3}$ for which orbital radius is $r_{1} .$
Put $Z = 4$ and $r_{n} = r_{1} = k$ in Eq.(1)
$r_{1} = \frac{r_{1} n^{2}}{4} \Rightarrow n = 2$

PHXII12:ATOMS

356508 As per Bohr model the minimum energy (in $e V$ ) required to remove the electron from the ground state of doubly ionised lithium $(Z = 3)$ is

1

1.51

2

13.6

3

40.8

4

122.4

Explanation:

Ionsiation energy
$= Z^{2} R h c (\frac{1}{{(1)}^{2}} - \frac{1}{{(\infty)}^{2}})$
$= Z^{2} R h c = {(3)}^{2} \times 13.6 e V = 122.4 e V$

PHXII12:ATOMS

356509 An electron of a hydrogen like atom, having $Z = 4$ , jumps from $4^{th}$ energy state to $2^{nd}$ energy state. The energy released in this process, will be (Given $R c h = 13.6 e V$ ) Where, $R =$ Rydberg constant, $c =$ Speed of light in vacuum, $h =$ Planck's constant

1

40.8 e V

2

13.6 e V

3

3.4 e V

4

10.5 e V

Explanation:

As, $Z = 4, n_{f} = 2, n_{i} = 4$
The energy released is given by
$Δ E = 13.6 Z^{2} [\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}] = 13.6 \times 4^{2} [\frac{1}{4} - \frac{1}{16}]$
$Δ E = 13.6 \times 16 [\frac{4 - 1}{16}]$
$= 13.6 \times 3 = 40.8 e V$

JEE - 2023

PHXII12:ATOMS

356510 For which one of the following, Bohr model is not valid?

1 Singly ionized helium atom

(H e^{+})

2 Deuteron atom

3 Singly ionized neon atom

(N e^{+})

4 Hydrogen atom

NEET - 2020

PHXII12:ATOMS

356511 Assertion :
There is a difference in the spectrum of sunlight when passed through "yellow-glass" and that of the light from a sodium vapour lamp.
Reason :
The sunlight always gives continuous spectrum with absorption lines (the Fraunhofer lines).

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII12:ATOMS

356512 Which energy state of the triply ionised beryllium has the same electron orbital radius as that of ground state of hydrogen? Given $Z$ for $B e = 4.$

1

n = 1

2

n = 2

3

n = 3

4

n = 4

Explanation:

For an atom of atomic number $Z$ , radius of $n^{t h}$ orbit is given by
$r_{n} = \frac{k n^{2}}{Z} (1)$
where $k = c o n s t a n t$
For ground state of hydrogen, $Z = 1, n = 1,$ so that $r_{1} = k$
Let $n$ be the energy state of $B e^{+ 3}$ for which orbital radius is $r_{1} .$
Put $Z = 4$ and $r_{n} = r_{1} = k$ in Eq.(1)
$r_{1} = \frac{r_{1} n^{2}}{4} \Rightarrow n = 2$

PHXII12:ATOMS

356508 As per Bohr model the minimum energy (in $e V$ ) required to remove the electron from the ground state of doubly ionised lithium $(Z = 3)$ is

1

1.51

2

13.6

3

40.8

4

122.4

Explanation:

Ionsiation energy
$= Z^{2} R h c (\frac{1}{{(1)}^{2}} - \frac{1}{{(\infty)}^{2}})$
$= Z^{2} R h c = {(3)}^{2} \times 13.6 e V = 122.4 e V$

PHXII12:ATOMS

356509 An electron of a hydrogen like atom, having $Z = 4$ , jumps from $4^{th}$ energy state to $2^{nd}$ energy state. The energy released in this process, will be (Given $R c h = 13.6 e V$ ) Where, $R =$ Rydberg constant, $c =$ Speed of light in vacuum, $h =$ Planck's constant

1

40.8 e V

2

13.6 e V

3

3.4 e V

4

10.5 e V

Explanation:

As, $Z = 4, n_{f} = 2, n_{i} = 4$
The energy released is given by
$Δ E = 13.6 Z^{2} [\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}] = 13.6 \times 4^{2} [\frac{1}{4} - \frac{1}{16}]$
$Δ E = 13.6 \times 16 [\frac{4 - 1}{16}]$
$= 13.6 \times 3 = 40.8 e V$

JEE - 2023

PHXII12:ATOMS

356510 For which one of the following, Bohr model is not valid?

1 Singly ionized helium atom

(H e^{+})

2 Deuteron atom

3 Singly ionized neon atom

(N e^{+})

4 Hydrogen atom

NEET - 2020

PHXII12:ATOMS

356511 Assertion :
There is a difference in the spectrum of sunlight when passed through "yellow-glass" and that of the light from a sodium vapour lamp.
Reason :
The sunlight always gives continuous spectrum with absorption lines (the Fraunhofer lines).

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII12:ATOMS

356512 Which energy state of the triply ionised beryllium has the same electron orbital radius as that of ground state of hydrogen? Given $Z$ for $B e = 4.$

1

n = 1

2

n = 2

3

n = 3

4

n = 4

Explanation:

For an atom of atomic number $Z$ , radius of $n^{t h}$ orbit is given by
$r_{n} = \frac{k n^{2}}{Z} (1)$
where $k = c o n s t a n t$
For ground state of hydrogen, $Z = 1, n = 1,$ so that $r_{1} = k$
Let $n$ be the energy state of $B e^{+ 3}$ for which orbital radius is $r_{1} .$
Put $Z = 4$ and $r_{n} = r_{1} = k$ in Eq.(1)
$r_{1} = \frac{r_{1} n^{2}}{4} \Rightarrow n = 2$

PHXII12:ATOMS

356508 As per Bohr model the minimum energy (in $e V$ ) required to remove the electron from the ground state of doubly ionised lithium $(Z = 3)$ is

1

1.51

2

13.6

3

40.8

4

122.4

Explanation:

Ionsiation energy
$= Z^{2} R h c (\frac{1}{{(1)}^{2}} - \frac{1}{{(\infty)}^{2}})$
$= Z^{2} R h c = {(3)}^{2} \times 13.6 e V = 122.4 e V$

PHXII12:ATOMS

356509 An electron of a hydrogen like atom, having $Z = 4$ , jumps from $4^{th}$ energy state to $2^{nd}$ energy state. The energy released in this process, will be (Given $R c h = 13.6 e V$ ) Where, $R =$ Rydberg constant, $c =$ Speed of light in vacuum, $h =$ Planck's constant

1

40.8 e V

2

13.6 e V

3

3.4 e V

4

10.5 e V

Explanation:

As, $Z = 4, n_{f} = 2, n_{i} = 4$
The energy released is given by
$Δ E = 13.6 Z^{2} [\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}] = 13.6 \times 4^{2} [\frac{1}{4} - \frac{1}{16}]$
$Δ E = 13.6 \times 16 [\frac{4 - 1}{16}]$
$= 13.6 \times 3 = 40.8 e V$

JEE - 2023

PHXII12:ATOMS

356510 For which one of the following, Bohr model is not valid?

1 Singly ionized helium atom

(H e^{+})

2 Deuteron atom

3 Singly ionized neon atom

(N e^{+})

4 Hydrogen atom

NEET - 2020

PHXII12:ATOMS

356511 Assertion :
There is a difference in the spectrum of sunlight when passed through "yellow-glass" and that of the light from a sodium vapour lamp.
Reason :
The sunlight always gives continuous spectrum with absorption lines (the Fraunhofer lines).

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII12:ATOMS

356512 Which energy state of the triply ionised beryllium has the same electron orbital radius as that of ground state of hydrogen? Given $Z$ for $B e = 4.$

1

n = 1

2

n = 2

3

n = 3

4

n = 4

Explanation:

For an atom of atomic number $Z$ , radius of $n^{t h}$ orbit is given by
$r_{n} = \frac{k n^{2}}{Z} (1)$
where $k = c o n s t a n t$
For ground state of hydrogen, $Z = 1, n = 1,$ so that $r_{1} = k$
Let $n$ be the energy state of $B e^{+ 3}$ for which orbital radius is $r_{1} .$
Put $Z = 4$ and $r_{n} = r_{1} = k$ in Eq.(1)
$r_{1} = \frac{r_{1} n^{2}}{4} \Rightarrow n = 2$

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