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ATOMS

145251 The energy levels of a hypothetical one electron atom system are given by
$E_{n} = - \frac{16}{n^{2}} e V, where n = 1, 2, 3, \dots .$
The wavelength of emitted photon corresponding to transition from first excited level to ground level is about

1

1035 \AA

2

1220 \AA

3

3650 \AA

4

690 \AA

Explanation:

A Given,
$E_{n} = \frac{- 16}{n^{2}} eV$
Where, $n = 1, 2, 3, \dots \dots$
First excited level means, $n = 2$ and ground level, $n = 1$ Then,
$E_{1} = \frac{- 16}{1} eV = - 16 eV$
$E_{2} = \frac{- 16}{2^{2}} eV = - 4 eV$
$Δ E = E_{2} - E_{1} = - 4 - (- 16) = 12 eV$
We know that,
$λ = \frac{12420}{Δ E} = \frac{12420}{12} = 1035 \AA$
$λ = 1035 \AA$

UPSEE - 2017

ATOMS

145256 What will be ratio of radii of ${Li}^{7}$ nucleus to ${Fe}^{56}$ nucleus?

1

1 : 3

2

1 : 2

3

1 : 8

4

2 : 6

Explanation:

B Given,
Mass of $Li (A) = 7$
Mass of $Fe (A) = 56$
We know that,
Radius of nucleus, $R = R_{0} (A)^{1 / 3}$
Where, $R_{0} =$ constant and value of $R_{0} = 1.2 \times 10^{- 15} m$
Radii of $Li = R_{0} (7)^{1 / 3}$
Radii of $Fe = R_{0} (56)^{1 / 3}$
$∵ \frac{R_{Li}}{R_{Fe}} = \frac{R_{0} (7)^{1 / 3}}{R_{0} (56)^{1 / 3}} = {(\frac{7}{56})}^{1 / 3}$
$\frac{R_{Li}}{R_{Fe}} = {(\frac{1}{8})}^{3} = \frac{1}{2}$
$R_{Li} : R_{Fe} = 1 : 2$

CG PET- 2008

ATOMS

145257 An electron is at ground state of the $H$ atom. Minimum energy required to excite the $H$ atom into second excited state is

1

13.6 eV

2

12.1 eV

3

10.2 eV

4

3.4 eV

Explanation:

B Hydrogen atom having only one electron, then the energy required to excite hydrogen atom will be equal to the energy required to excite one electron to second level,
$E = \frac{- 13.6 Z^{2}}{n^{2}}$
Where, $Z$ is atomic number and $n$ is energy state.
For $Z = 1, n = 1$
$E_{1} = \frac{- 13.6 \times 1^{2}}{1^{2}} = - 13.6 eV$
For an electron jump from first excited state to second.
$∴ Z = 1, n = 3$
$E_{3} = \frac{- 13.6 \times 1^{2}}{3^{2}} = - 1.5 eV$
Therefore, $Δ E = E_{3} - E_{1}$
$Δ E = - 1.5 - (- 13.6)$
$Δ E = 12.1 eV$

UPSEE - 2016

ATOMS

145261 The radius of germanium (Ge) nuclide is measured to be twice the radius of $_{4}^{9} Be$ . The number of nucleons in Ge are :

1 73

2 74

3 75

4 72

Explanation:

D Given,
Radius of germanium $= 2$ times radius of Beryllium
$R_{Ge} = 2 R_{Be}$
Mass of Beryllium (A) $= 9$
Mass of germanium $(A) =$ ?
We know that,
Radius of nucleus, $R = R_{0} (A)^{1 / 3}$ Where, $R_{0} =$ Constant and value of $R_{0} = 1.2 \times 10^{- 15} m$
Or $R \propto (A)^{1 / 3}$
$∴ \frac{R_{Be}}{R_{Ge}} = \frac{(9)^{1 / 3}}{(A)^{1 / 3}}$
From equation (i), we get-
$∴ \frac{R_{Be}}{2 R_{Be}} = \frac{(9)^{1 / 3}}{(A)^{1 / 3}}$
$(A)^{1 / 3} = 2 \times (9)^{1 / 3}$
$A = 2^{3} \times 9$
$∴$ The number of nucleons in $Ge$ is 72 .

UPSEE - 2006

ATOMS

145251 The energy levels of a hypothetical one electron atom system are given by
$E_{n} = - \frac{16}{n^{2}} e V, where n = 1, 2, 3, \dots .$
The wavelength of emitted photon corresponding to transition from first excited level to ground level is about

1

1035 \AA

2

1220 \AA

3

3650 \AA

4

690 \AA

Explanation:

A Given,
$E_{n} = \frac{- 16}{n^{2}} eV$
Where, $n = 1, 2, 3, \dots \dots$
First excited level means, $n = 2$ and ground level, $n = 1$ Then,
$E_{1} = \frac{- 16}{1} eV = - 16 eV$
$E_{2} = \frac{- 16}{2^{2}} eV = - 4 eV$
$Δ E = E_{2} - E_{1} = - 4 - (- 16) = 12 eV$
We know that,
$λ = \frac{12420}{Δ E} = \frac{12420}{12} = 1035 \AA$
$λ = 1035 \AA$

UPSEE - 2017

ATOMS

145256 What will be ratio of radii of ${Li}^{7}$ nucleus to ${Fe}^{56}$ nucleus?

1

1 : 3

2

1 : 2

3

1 : 8

4

2 : 6

Explanation:

B Given,
Mass of $Li (A) = 7$
Mass of $Fe (A) = 56$
We know that,
Radius of nucleus, $R = R_{0} (A)^{1 / 3}$
Where, $R_{0} =$ constant and value of $R_{0} = 1.2 \times 10^{- 15} m$
Radii of $Li = R_{0} (7)^{1 / 3}$
Radii of $Fe = R_{0} (56)^{1 / 3}$
$∵ \frac{R_{Li}}{R_{Fe}} = \frac{R_{0} (7)^{1 / 3}}{R_{0} (56)^{1 / 3}} = {(\frac{7}{56})}^{1 / 3}$
$\frac{R_{Li}}{R_{Fe}} = {(\frac{1}{8})}^{3} = \frac{1}{2}$
$R_{Li} : R_{Fe} = 1 : 2$

CG PET- 2008

ATOMS

145257 An electron is at ground state of the $H$ atom. Minimum energy required to excite the $H$ atom into second excited state is

1

13.6 eV

2

12.1 eV

3

10.2 eV

4

3.4 eV

Explanation:

B Hydrogen atom having only one electron, then the energy required to excite hydrogen atom will be equal to the energy required to excite one electron to second level,
$E = \frac{- 13.6 Z^{2}}{n^{2}}$
Where, $Z$ is atomic number and $n$ is energy state.
For $Z = 1, n = 1$
$E_{1} = \frac{- 13.6 \times 1^{2}}{1^{2}} = - 13.6 eV$
For an electron jump from first excited state to second.
$∴ Z = 1, n = 3$
$E_{3} = \frac{- 13.6 \times 1^{2}}{3^{2}} = - 1.5 eV$
Therefore, $Δ E = E_{3} - E_{1}$
$Δ E = - 1.5 - (- 13.6)$
$Δ E = 12.1 eV$

UPSEE - 2016

ATOMS

145261 The radius of germanium (Ge) nuclide is measured to be twice the radius of $_{4}^{9} Be$ . The number of nucleons in Ge are :

1 73

2 74

3 75

4 72

Explanation:

D Given,
Radius of germanium $= 2$ times radius of Beryllium
$R_{Ge} = 2 R_{Be}$
Mass of Beryllium (A) $= 9$
Mass of germanium $(A) =$ ?
We know that,
Radius of nucleus, $R = R_{0} (A)^{1 / 3}$ Where, $R_{0} =$ Constant and value of $R_{0} = 1.2 \times 10^{- 15} m$
Or $R \propto (A)^{1 / 3}$
$∴ \frac{R_{Be}}{R_{Ge}} = \frac{(9)^{1 / 3}}{(A)^{1 / 3}}$
From equation (i), we get-
$∴ \frac{R_{Be}}{2 R_{Be}} = \frac{(9)^{1 / 3}}{(A)^{1 / 3}}$
$(A)^{1 / 3} = 2 \times (9)^{1 / 3}$
$A = 2^{3} \times 9$
$∴$ The number of nucleons in $Ge$ is 72 .

UPSEE - 2006

ATOMS

145251 The energy levels of a hypothetical one electron atom system are given by
$E_{n} = - \frac{16}{n^{2}} e V, where n = 1, 2, 3, \dots .$
The wavelength of emitted photon corresponding to transition from first excited level to ground level is about

1

1035 \AA

2

1220 \AA

3

3650 \AA

4

690 \AA

Explanation:

A Given,
$E_{n} = \frac{- 16}{n^{2}} eV$
Where, $n = 1, 2, 3, \dots \dots$
First excited level means, $n = 2$ and ground level, $n = 1$ Then,
$E_{1} = \frac{- 16}{1} eV = - 16 eV$
$E_{2} = \frac{- 16}{2^{2}} eV = - 4 eV$
$Δ E = E_{2} - E_{1} = - 4 - (- 16) = 12 eV$
We know that,
$λ = \frac{12420}{Δ E} = \frac{12420}{12} = 1035 \AA$
$λ = 1035 \AA$

UPSEE - 2017

ATOMS

145256 What will be ratio of radii of ${Li}^{7}$ nucleus to ${Fe}^{56}$ nucleus?

1

1 : 3

2

1 : 2

3

1 : 8

4

2 : 6

Explanation:

B Given,
Mass of $Li (A) = 7$
Mass of $Fe (A) = 56$
We know that,
Radius of nucleus, $R = R_{0} (A)^{1 / 3}$
Where, $R_{0} =$ constant and value of $R_{0} = 1.2 \times 10^{- 15} m$
Radii of $Li = R_{0} (7)^{1 / 3}$
Radii of $Fe = R_{0} (56)^{1 / 3}$
$∵ \frac{R_{Li}}{R_{Fe}} = \frac{R_{0} (7)^{1 / 3}}{R_{0} (56)^{1 / 3}} = {(\frac{7}{56})}^{1 / 3}$
$\frac{R_{Li}}{R_{Fe}} = {(\frac{1}{8})}^{3} = \frac{1}{2}$
$R_{Li} : R_{Fe} = 1 : 2$

CG PET- 2008

ATOMS

145257 An electron is at ground state of the $H$ atom. Minimum energy required to excite the $H$ atom into second excited state is

1

13.6 eV

2

12.1 eV

3

10.2 eV

4

3.4 eV

Explanation:

B Hydrogen atom having only one electron, then the energy required to excite hydrogen atom will be equal to the energy required to excite one electron to second level,
$E = \frac{- 13.6 Z^{2}}{n^{2}}$
Where, $Z$ is atomic number and $n$ is energy state.
For $Z = 1, n = 1$
$E_{1} = \frac{- 13.6 \times 1^{2}}{1^{2}} = - 13.6 eV$
For an electron jump from first excited state to second.
$∴ Z = 1, n = 3$
$E_{3} = \frac{- 13.6 \times 1^{2}}{3^{2}} = - 1.5 eV$
Therefore, $Δ E = E_{3} - E_{1}$
$Δ E = - 1.5 - (- 13.6)$
$Δ E = 12.1 eV$

UPSEE - 2016

ATOMS

145261 The radius of germanium (Ge) nuclide is measured to be twice the radius of $_{4}^{9} Be$ . The number of nucleons in Ge are :

1 73

2 74

3 75

4 72

Explanation:

D Given,
Radius of germanium $= 2$ times radius of Beryllium
$R_{Ge} = 2 R_{Be}$
Mass of Beryllium (A) $= 9$
Mass of germanium $(A) =$ ?
We know that,
Radius of nucleus, $R = R_{0} (A)^{1 / 3}$ Where, $R_{0} =$ Constant and value of $R_{0} = 1.2 \times 10^{- 15} m$
Or $R \propto (A)^{1 / 3}$
$∴ \frac{R_{Be}}{R_{Ge}} = \frac{(9)^{1 / 3}}{(A)^{1 / 3}}$
From equation (i), we get-
$∴ \frac{R_{Be}}{2 R_{Be}} = \frac{(9)^{1 / 3}}{(A)^{1 / 3}}$
$(A)^{1 / 3} = 2 \times (9)^{1 / 3}$
$A = 2^{3} \times 9$
$∴$ The number of nucleons in $Ge$ is 72 .

UPSEE - 2006

ATOMS

145251 The energy levels of a hypothetical one electron atom system are given by
$E_{n} = - \frac{16}{n^{2}} e V, where n = 1, 2, 3, \dots .$
The wavelength of emitted photon corresponding to transition from first excited level to ground level is about

1

1035 \AA

2

1220 \AA

3

3650 \AA

4

690 \AA

Explanation:

A Given,
$E_{n} = \frac{- 16}{n^{2}} eV$
Where, $n = 1, 2, 3, \dots \dots$
First excited level means, $n = 2$ and ground level, $n = 1$ Then,
$E_{1} = \frac{- 16}{1} eV = - 16 eV$
$E_{2} = \frac{- 16}{2^{2}} eV = - 4 eV$
$Δ E = E_{2} - E_{1} = - 4 - (- 16) = 12 eV$
We know that,
$λ = \frac{12420}{Δ E} = \frac{12420}{12} = 1035 \AA$
$λ = 1035 \AA$

UPSEE - 2017

ATOMS

145256 What will be ratio of radii of ${Li}^{7}$ nucleus to ${Fe}^{56}$ nucleus?

1

1 : 3

2

1 : 2

3

1 : 8

4

2 : 6

Explanation:

B Given,
Mass of $Li (A) = 7$
Mass of $Fe (A) = 56$
We know that,
Radius of nucleus, $R = R_{0} (A)^{1 / 3}$
Where, $R_{0} =$ constant and value of $R_{0} = 1.2 \times 10^{- 15} m$
Radii of $Li = R_{0} (7)^{1 / 3}$
Radii of $Fe = R_{0} (56)^{1 / 3}$
$∵ \frac{R_{Li}}{R_{Fe}} = \frac{R_{0} (7)^{1 / 3}}{R_{0} (56)^{1 / 3}} = {(\frac{7}{56})}^{1 / 3}$
$\frac{R_{Li}}{R_{Fe}} = {(\frac{1}{8})}^{3} = \frac{1}{2}$
$R_{Li} : R_{Fe} = 1 : 2$

CG PET- 2008

ATOMS

145257 An electron is at ground state of the $H$ atom. Minimum energy required to excite the $H$ atom into second excited state is

1

13.6 eV

2

12.1 eV

3

10.2 eV

4

3.4 eV

Explanation:

B Hydrogen atom having only one electron, then the energy required to excite hydrogen atom will be equal to the energy required to excite one electron to second level,
$E = \frac{- 13.6 Z^{2}}{n^{2}}$
Where, $Z$ is atomic number and $n$ is energy state.
For $Z = 1, n = 1$
$E_{1} = \frac{- 13.6 \times 1^{2}}{1^{2}} = - 13.6 eV$
For an electron jump from first excited state to second.
$∴ Z = 1, n = 3$
$E_{3} = \frac{- 13.6 \times 1^{2}}{3^{2}} = - 1.5 eV$
Therefore, $Δ E = E_{3} - E_{1}$
$Δ E = - 1.5 - (- 13.6)$
$Δ E = 12.1 eV$

UPSEE - 2016

ATOMS

145261 The radius of germanium (Ge) nuclide is measured to be twice the radius of $_{4}^{9} Be$ . The number of nucleons in Ge are :

1 73

2 74

3 75

4 72

Explanation:

D Given,
Radius of germanium $= 2$ times radius of Beryllium
$R_{Ge} = 2 R_{Be}$
Mass of Beryllium (A) $= 9$
Mass of germanium $(A) =$ ?
We know that,
Radius of nucleus, $R = R_{0} (A)^{1 / 3}$ Where, $R_{0} =$ Constant and value of $R_{0} = 1.2 \times 10^{- 15} m$
Or $R \propto (A)^{1 / 3}$
$∴ \frac{R_{Be}}{R_{Ge}} = \frac{(9)^{1 / 3}}{(A)^{1 / 3}}$
From equation (i), we get-
$∴ \frac{R_{Be}}{2 R_{Be}} = \frac{(9)^{1 / 3}}{(A)^{1 / 3}}$
$(A)^{1 / 3} = 2 \times (9)^{1 / 3}$
$A = 2^{3} \times 9$
$∴$ The number of nucleons in $Ge$ is 72 .

UPSEE - 2006