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CHXI02:STRUCTURE OF ATOM

307146 The maximum energy is possessed by an electron, when it is present

1 in nucleus

2 in ground energy state

3 in first excited state

4 at infinite distance from the nucleus

Explanation:

As $E_{n} = - 21.8 \times 10^{- 19} \times Z^{2} / n^{2} J / a t o m$
Thus, $E_{n} \propto [- \frac{1}{n^{2}}]$
where $n$ is the number of orbits.
Therefore, as the value of n increases, energy of the electron also increases. Therefore, when $n$ becomes infinite, energy also becomes infinite. Due to this reason, maximum energy is possessed by an electron, when it is present at infinite distance from the nucleus.

CHXI02:STRUCTURE OF ATOM

307147 What is the energy associated with first orbit of ${Li}^{2 +} (R_{H} = 2.18 \times 10^{- 18} J) ?$

1

- 19.62 \times 10^{- 18} J

2

- 34.88 \times 10^{- 18} J

3

- 2.18 \times 10^{- 18} J

4

- 8.72 \times 10^{- 18} J

Explanation:

$E_{n} = - R_{H} \frac{Z^{2}}{n^{2}} = \frac{- 2.18 \times 10^{- 18} \times Z^{2}}{n^{2}} J$
For ${Li}^{2 +}; n = 1, Z = 3$
$E_{n} = \frac{- 2 .18 \times 1 0^{- 18} \times 3^{2}}{1^{2}}$
$J = - 2 .18 \times 1 0^{- 18} \times 9J$
$= - 19 .62 \times 1 0^{- 18} J$

MHTCET - 2022

CHXI02:STRUCTURE OF ATOM

307159 The ionization enthalpy of hydrogen atom is $1 .312 \times 1 0^{6} J m o l^{- 1} .$ The energy required to excite the electron in the atom from $n = 1$ to $n = 2$ is

1

9 .84 \times 1 0^{5} J m o l^{- 1}

2

8 .50 \times 1 0^{5} J m o l^{- 1}

3

6 .56 \times 1 0^{5} J m o l^{- 1}

4

7 .56 \times 1 0^{5} J m o l^{- 1}

Explanation:

$E_{1} = - 1 .312 \times 1 0^{6} J m o l^{- 1}$
$E_{2} = \frac{- 1 .312 \times 1 0^{6}}{2^{2}} J m o l^{- 1}$
$Δ E = E_{2} - E_{1} = \frac{- 1 .312 \times 1 0^{6}}{4} - (- 1 .312 \times 1 0^{6})$
$= 9 .84 \times 1 0^{5} J m o l^{- 1}$

CHXI02:STRUCTURE OF ATOM

307148 Energy of an electron is given by $E = - 2 .178 \times 1 0^{- 18} J (\frac{Z^{2}}{n^{2}})$ . Wavelength of light required to excite an electron in an hydrogen atom from level $n = 1$ to $n = 2$ will be $(h = 6 .62 \times 1 0^{- 34} J . s a n d c = 3 .0 \times 1 0^{8} m / s)$

1

1 .214 \times 1 0^{- 7} m

2

2 .816 \times 1 0^{- 17} m

3

6 .500 \times 1 0^{- 17} m

4

8 .500 \times 1 0^{- 7} m

Explanation:

Energy change, $Δ E = E_{2} - E_{1}$
$= 2 .178 \times 1 0^{- 18} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] J$
Energy change, $Δ E = 2 .178 \times 1 0^{- 18} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = \frac{h c}{λ}$
$2 .178 \times 1 0^{- 18} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = \frac{6 .62 \times 1 0^{- 34} \times 3 .0 \times 1 0^{8}}{λ}$
$∴ λ = 1 .214 \times 1 0^{- 7} m$

CHXI02:STRUCTURE OF ATOM

307146 The maximum energy is possessed by an electron, when it is present

1 in nucleus

2 in ground energy state

3 in first excited state

4 at infinite distance from the nucleus

Explanation:

As $E_{n} = - 21.8 \times 10^{- 19} \times Z^{2} / n^{2} J / a t o m$
Thus, $E_{n} \propto [- \frac{1}{n^{2}}]$
where $n$ is the number of orbits.
Therefore, as the value of n increases, energy of the electron also increases. Therefore, when $n$ becomes infinite, energy also becomes infinite. Due to this reason, maximum energy is possessed by an electron, when it is present at infinite distance from the nucleus.

CHXI02:STRUCTURE OF ATOM

307147 What is the energy associated with first orbit of ${Li}^{2 +} (R_{H} = 2.18 \times 10^{- 18} J) ?$

1

- 19.62 \times 10^{- 18} J

2

- 34.88 \times 10^{- 18} J

3

- 2.18 \times 10^{- 18} J

4

- 8.72 \times 10^{- 18} J

Explanation:

$E_{n} = - R_{H} \frac{Z^{2}}{n^{2}} = \frac{- 2.18 \times 10^{- 18} \times Z^{2}}{n^{2}} J$
For ${Li}^{2 +}; n = 1, Z = 3$
$E_{n} = \frac{- 2 .18 \times 1 0^{- 18} \times 3^{2}}{1^{2}}$
$J = - 2 .18 \times 1 0^{- 18} \times 9J$
$= - 19 .62 \times 1 0^{- 18} J$

MHTCET - 2022

CHXI02:STRUCTURE OF ATOM

307159 The ionization enthalpy of hydrogen atom is $1 .312 \times 1 0^{6} J m o l^{- 1} .$ The energy required to excite the electron in the atom from $n = 1$ to $n = 2$ is

1

9 .84 \times 1 0^{5} J m o l^{- 1}

2

8 .50 \times 1 0^{5} J m o l^{- 1}

3

6 .56 \times 1 0^{5} J m o l^{- 1}

4

7 .56 \times 1 0^{5} J m o l^{- 1}

Explanation:

$E_{1} = - 1 .312 \times 1 0^{6} J m o l^{- 1}$
$E_{2} = \frac{- 1 .312 \times 1 0^{6}}{2^{2}} J m o l^{- 1}$
$Δ E = E_{2} - E_{1} = \frac{- 1 .312 \times 1 0^{6}}{4} - (- 1 .312 \times 1 0^{6})$
$= 9 .84 \times 1 0^{5} J m o l^{- 1}$

CHXI02:STRUCTURE OF ATOM

307148 Energy of an electron is given by $E = - 2 .178 \times 1 0^{- 18} J (\frac{Z^{2}}{n^{2}})$ . Wavelength of light required to excite an electron in an hydrogen atom from level $n = 1$ to $n = 2$ will be $(h = 6 .62 \times 1 0^{- 34} J . s a n d c = 3 .0 \times 1 0^{8} m / s)$

1

1 .214 \times 1 0^{- 7} m

2

2 .816 \times 1 0^{- 17} m

3

6 .500 \times 1 0^{- 17} m

4

8 .500 \times 1 0^{- 7} m

Explanation:

Energy change, $Δ E = E_{2} - E_{1}$
$= 2 .178 \times 1 0^{- 18} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] J$
Energy change, $Δ E = 2 .178 \times 1 0^{- 18} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = \frac{h c}{λ}$
$2 .178 \times 1 0^{- 18} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = \frac{6 .62 \times 1 0^{- 34} \times 3 .0 \times 1 0^{8}}{λ}$
$∴ λ = 1 .214 \times 1 0^{- 7} m$

CHXI02:STRUCTURE OF ATOM

307146 The maximum energy is possessed by an electron, when it is present

1 in nucleus

2 in ground energy state

3 in first excited state

4 at infinite distance from the nucleus

Explanation:

As $E_{n} = - 21.8 \times 10^{- 19} \times Z^{2} / n^{2} J / a t o m$
Thus, $E_{n} \propto [- \frac{1}{n^{2}}]$
where $n$ is the number of orbits.
Therefore, as the value of n increases, energy of the electron also increases. Therefore, when $n$ becomes infinite, energy also becomes infinite. Due to this reason, maximum energy is possessed by an electron, when it is present at infinite distance from the nucleus.

CHXI02:STRUCTURE OF ATOM

307147 What is the energy associated with first orbit of ${Li}^{2 +} (R_{H} = 2.18 \times 10^{- 18} J) ?$

1

- 19.62 \times 10^{- 18} J

2

- 34.88 \times 10^{- 18} J

3

- 2.18 \times 10^{- 18} J

4

- 8.72 \times 10^{- 18} J

Explanation:

$E_{n} = - R_{H} \frac{Z^{2}}{n^{2}} = \frac{- 2.18 \times 10^{- 18} \times Z^{2}}{n^{2}} J$
For ${Li}^{2 +}; n = 1, Z = 3$
$E_{n} = \frac{- 2 .18 \times 1 0^{- 18} \times 3^{2}}{1^{2}}$
$J = - 2 .18 \times 1 0^{- 18} \times 9J$
$= - 19 .62 \times 1 0^{- 18} J$

MHTCET - 2022

CHXI02:STRUCTURE OF ATOM

307159 The ionization enthalpy of hydrogen atom is $1 .312 \times 1 0^{6} J m o l^{- 1} .$ The energy required to excite the electron in the atom from $n = 1$ to $n = 2$ is

1

9 .84 \times 1 0^{5} J m o l^{- 1}

2

8 .50 \times 1 0^{5} J m o l^{- 1}

3

6 .56 \times 1 0^{5} J m o l^{- 1}

4

7 .56 \times 1 0^{5} J m o l^{- 1}

Explanation:

$E_{1} = - 1 .312 \times 1 0^{6} J m o l^{- 1}$
$E_{2} = \frac{- 1 .312 \times 1 0^{6}}{2^{2}} J m o l^{- 1}$
$Δ E = E_{2} - E_{1} = \frac{- 1 .312 \times 1 0^{6}}{4} - (- 1 .312 \times 1 0^{6})$
$= 9 .84 \times 1 0^{5} J m o l^{- 1}$

CHXI02:STRUCTURE OF ATOM

307148 Energy of an electron is given by $E = - 2 .178 \times 1 0^{- 18} J (\frac{Z^{2}}{n^{2}})$ . Wavelength of light required to excite an electron in an hydrogen atom from level $n = 1$ to $n = 2$ will be $(h = 6 .62 \times 1 0^{- 34} J . s a n d c = 3 .0 \times 1 0^{8} m / s)$

1

1 .214 \times 1 0^{- 7} m

2

2 .816 \times 1 0^{- 17} m

3

6 .500 \times 1 0^{- 17} m

4

8 .500 \times 1 0^{- 7} m

Explanation:

Energy change, $Δ E = E_{2} - E_{1}$
$= 2 .178 \times 1 0^{- 18} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] J$
Energy change, $Δ E = 2 .178 \times 1 0^{- 18} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = \frac{h c}{λ}$
$2 .178 \times 1 0^{- 18} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = \frac{6 .62 \times 1 0^{- 34} \times 3 .0 \times 1 0^{8}}{λ}$
$∴ λ = 1 .214 \times 1 0^{- 7} m$

CHXI02:STRUCTURE OF ATOM

307146 The maximum energy is possessed by an electron, when it is present

1 in nucleus

2 in ground energy state

3 in first excited state

4 at infinite distance from the nucleus

Explanation:

As $E_{n} = - 21.8 \times 10^{- 19} \times Z^{2} / n^{2} J / a t o m$
Thus, $E_{n} \propto [- \frac{1}{n^{2}}]$
where $n$ is the number of orbits.
Therefore, as the value of n increases, energy of the electron also increases. Therefore, when $n$ becomes infinite, energy also becomes infinite. Due to this reason, maximum energy is possessed by an electron, when it is present at infinite distance from the nucleus.

CHXI02:STRUCTURE OF ATOM

307147 What is the energy associated with first orbit of ${Li}^{2 +} (R_{H} = 2.18 \times 10^{- 18} J) ?$

1

- 19.62 \times 10^{- 18} J

2

- 34.88 \times 10^{- 18} J

3

- 2.18 \times 10^{- 18} J

4

- 8.72 \times 10^{- 18} J

Explanation:

$E_{n} = - R_{H} \frac{Z^{2}}{n^{2}} = \frac{- 2.18 \times 10^{- 18} \times Z^{2}}{n^{2}} J$
For ${Li}^{2 +}; n = 1, Z = 3$
$E_{n} = \frac{- 2 .18 \times 1 0^{- 18} \times 3^{2}}{1^{2}}$
$J = - 2 .18 \times 1 0^{- 18} \times 9J$
$= - 19 .62 \times 1 0^{- 18} J$

MHTCET - 2022

CHXI02:STRUCTURE OF ATOM

307159 The ionization enthalpy of hydrogen atom is $1 .312 \times 1 0^{6} J m o l^{- 1} .$ The energy required to excite the electron in the atom from $n = 1$ to $n = 2$ is

1

9 .84 \times 1 0^{5} J m o l^{- 1}

2

8 .50 \times 1 0^{5} J m o l^{- 1}

3

6 .56 \times 1 0^{5} J m o l^{- 1}

4

7 .56 \times 1 0^{5} J m o l^{- 1}

Explanation:

$E_{1} = - 1 .312 \times 1 0^{6} J m o l^{- 1}$
$E_{2} = \frac{- 1 .312 \times 1 0^{6}}{2^{2}} J m o l^{- 1}$
$Δ E = E_{2} - E_{1} = \frac{- 1 .312 \times 1 0^{6}}{4} - (- 1 .312 \times 1 0^{6})$
$= 9 .84 \times 1 0^{5} J m o l^{- 1}$

CHXI02:STRUCTURE OF ATOM

307148 Energy of an electron is given by $E = - 2 .178 \times 1 0^{- 18} J (\frac{Z^{2}}{n^{2}})$ . Wavelength of light required to excite an electron in an hydrogen atom from level $n = 1$ to $n = 2$ will be $(h = 6 .62 \times 1 0^{- 34} J . s a n d c = 3 .0 \times 1 0^{8} m / s)$

1

1 .214 \times 1 0^{- 7} m

2

2 .816 \times 1 0^{- 17} m

3

6 .500 \times 1 0^{- 17} m

4

8 .500 \times 1 0^{- 7} m

Explanation:

Energy change, $Δ E = E_{2} - E_{1}$
$= 2 .178 \times 1 0^{- 18} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] J$
Energy change, $Δ E = 2 .178 \times 1 0^{- 18} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = \frac{h c}{λ}$
$2 .178 \times 1 0^{- 18} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = \frac{6 .62 \times 1 0^{- 34} \times 3 .0 \times 1 0^{8}}{λ}$
$∴ λ = 1 .214 \times 1 0^{- 7} m$

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