356521
The ionisation energy of \(L{i^{2 + }}\) is equal to
1 \(9hcR\)
2 \(6hcR\)
3 \(2hcR\)
4 \(hcR\)
Explanation:
As, ionisation energy \( = Rch{Z^2}\) For lithium \(Z = 3\) \(\therefore \) Ionisation energy \( = {(3)^2}Rch = 9Rch\)
PHXII12:ATOMS
356522
If the energy of hydrogen atom in nth orbit \({E_n}\), then energy in the nth orbit of a single ionized helium atom will be:
1 \(4{E_n}\)
2 \({E_n}/4\)
3 \(2{E_n}\)
4 \({E_n}/2\)
Explanation:
The energy of the electron in the \({n^{th}}\) orbit is \(E\,\alpha \,{Z^2}\) Given,\({Z_H} = 1,{Z_{He}} = 2\) \(\frac{{{E_H}}}{{{E_{He}}}} = \frac{{Z_H^2}}{{Z_{He}^2}} = \frac{1}{4} \Rightarrow {E_{He}} = 4{E_H}\) Given,\({E_H} = {E_n} \Rightarrow {E_{He}} = 4{E_n}\)
PHXII12:ATOMS
356523
The de-Broglie wavelength of an electron in 4th orbit is (\(r = \) radius of 1st orbit)
1 \(2\pi r\)
2 \(4\pi r\)
3 \(8\pi r\)
4 \(16\pi r\)
Explanation:
According to de Broglie’s quantisation condition, \(2\pi r = n\lambda \) \( \Rightarrow \lambda = \frac{{2\pi r}}{n}\) \(r\,\alpha \,{n^2}\) \(\frac{{{r_1}}}{{{r_2}}} = {\left( {\frac{1}{4}} \right)^2} \Rightarrow {r_2} = 16{r_1}\) Hence, \(\lambda = \frac{{2\pi [16{r_1}]}}{4} = 8\pi {r_1} = 8\pi r\)
PHXII12:ATOMS
356524
Assertion : Smoky flame of Bunsen burner gives continuous spectrum whereas its blue flame gives band spectrum. Reason : The band spectrum consists of coloured bands of light on a dark background.
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.
Explanation:
When the flame of a Bunsen burner appears smoky, it indicates that there are carbon particles within it that are in an incandescent state. As a result, the flame emits (like grey body)a continuous spectrum of light. However, when the burner produces a blue flame, it signifies the presence of carbon and cyanogen, in a molecular state. This type of flame yields a band spectrum of light. So correct option is (2).
356521
The ionisation energy of \(L{i^{2 + }}\) is equal to
1 \(9hcR\)
2 \(6hcR\)
3 \(2hcR\)
4 \(hcR\)
Explanation:
As, ionisation energy \( = Rch{Z^2}\) For lithium \(Z = 3\) \(\therefore \) Ionisation energy \( = {(3)^2}Rch = 9Rch\)
PHXII12:ATOMS
356522
If the energy of hydrogen atom in nth orbit \({E_n}\), then energy in the nth orbit of a single ionized helium atom will be:
1 \(4{E_n}\)
2 \({E_n}/4\)
3 \(2{E_n}\)
4 \({E_n}/2\)
Explanation:
The energy of the electron in the \({n^{th}}\) orbit is \(E\,\alpha \,{Z^2}\) Given,\({Z_H} = 1,{Z_{He}} = 2\) \(\frac{{{E_H}}}{{{E_{He}}}} = \frac{{Z_H^2}}{{Z_{He}^2}} = \frac{1}{4} \Rightarrow {E_{He}} = 4{E_H}\) Given,\({E_H} = {E_n} \Rightarrow {E_{He}} = 4{E_n}\)
PHXII12:ATOMS
356523
The de-Broglie wavelength of an electron in 4th orbit is (\(r = \) radius of 1st orbit)
1 \(2\pi r\)
2 \(4\pi r\)
3 \(8\pi r\)
4 \(16\pi r\)
Explanation:
According to de Broglie’s quantisation condition, \(2\pi r = n\lambda \) \( \Rightarrow \lambda = \frac{{2\pi r}}{n}\) \(r\,\alpha \,{n^2}\) \(\frac{{{r_1}}}{{{r_2}}} = {\left( {\frac{1}{4}} \right)^2} \Rightarrow {r_2} = 16{r_1}\) Hence, \(\lambda = \frac{{2\pi [16{r_1}]}}{4} = 8\pi {r_1} = 8\pi r\)
PHXII12:ATOMS
356524
Assertion : Smoky flame of Bunsen burner gives continuous spectrum whereas its blue flame gives band spectrum. Reason : The band spectrum consists of coloured bands of light on a dark background.
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.
Explanation:
When the flame of a Bunsen burner appears smoky, it indicates that there are carbon particles within it that are in an incandescent state. As a result, the flame emits (like grey body)a continuous spectrum of light. However, when the burner produces a blue flame, it signifies the presence of carbon and cyanogen, in a molecular state. This type of flame yields a band spectrum of light. So correct option is (2).
356521
The ionisation energy of \(L{i^{2 + }}\) is equal to
1 \(9hcR\)
2 \(6hcR\)
3 \(2hcR\)
4 \(hcR\)
Explanation:
As, ionisation energy \( = Rch{Z^2}\) For lithium \(Z = 3\) \(\therefore \) Ionisation energy \( = {(3)^2}Rch = 9Rch\)
PHXII12:ATOMS
356522
If the energy of hydrogen atom in nth orbit \({E_n}\), then energy in the nth orbit of a single ionized helium atom will be:
1 \(4{E_n}\)
2 \({E_n}/4\)
3 \(2{E_n}\)
4 \({E_n}/2\)
Explanation:
The energy of the electron in the \({n^{th}}\) orbit is \(E\,\alpha \,{Z^2}\) Given,\({Z_H} = 1,{Z_{He}} = 2\) \(\frac{{{E_H}}}{{{E_{He}}}} = \frac{{Z_H^2}}{{Z_{He}^2}} = \frac{1}{4} \Rightarrow {E_{He}} = 4{E_H}\) Given,\({E_H} = {E_n} \Rightarrow {E_{He}} = 4{E_n}\)
PHXII12:ATOMS
356523
The de-Broglie wavelength of an electron in 4th orbit is (\(r = \) radius of 1st orbit)
1 \(2\pi r\)
2 \(4\pi r\)
3 \(8\pi r\)
4 \(16\pi r\)
Explanation:
According to de Broglie’s quantisation condition, \(2\pi r = n\lambda \) \( \Rightarrow \lambda = \frac{{2\pi r}}{n}\) \(r\,\alpha \,{n^2}\) \(\frac{{{r_1}}}{{{r_2}}} = {\left( {\frac{1}{4}} \right)^2} \Rightarrow {r_2} = 16{r_1}\) Hence, \(\lambda = \frac{{2\pi [16{r_1}]}}{4} = 8\pi {r_1} = 8\pi r\)
PHXII12:ATOMS
356524
Assertion : Smoky flame of Bunsen burner gives continuous spectrum whereas its blue flame gives band spectrum. Reason : The band spectrum consists of coloured bands of light on a dark background.
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.
Explanation:
When the flame of a Bunsen burner appears smoky, it indicates that there are carbon particles within it that are in an incandescent state. As a result, the flame emits (like grey body)a continuous spectrum of light. However, when the burner produces a blue flame, it signifies the presence of carbon and cyanogen, in a molecular state. This type of flame yields a band spectrum of light. So correct option is (2).
356521
The ionisation energy of \(L{i^{2 + }}\) is equal to
1 \(9hcR\)
2 \(6hcR\)
3 \(2hcR\)
4 \(hcR\)
Explanation:
As, ionisation energy \( = Rch{Z^2}\) For lithium \(Z = 3\) \(\therefore \) Ionisation energy \( = {(3)^2}Rch = 9Rch\)
PHXII12:ATOMS
356522
If the energy of hydrogen atom in nth orbit \({E_n}\), then energy in the nth orbit of a single ionized helium atom will be:
1 \(4{E_n}\)
2 \({E_n}/4\)
3 \(2{E_n}\)
4 \({E_n}/2\)
Explanation:
The energy of the electron in the \({n^{th}}\) orbit is \(E\,\alpha \,{Z^2}\) Given,\({Z_H} = 1,{Z_{He}} = 2\) \(\frac{{{E_H}}}{{{E_{He}}}} = \frac{{Z_H^2}}{{Z_{He}^2}} = \frac{1}{4} \Rightarrow {E_{He}} = 4{E_H}\) Given,\({E_H} = {E_n} \Rightarrow {E_{He}} = 4{E_n}\)
PHXII12:ATOMS
356523
The de-Broglie wavelength of an electron in 4th orbit is (\(r = \) radius of 1st orbit)
1 \(2\pi r\)
2 \(4\pi r\)
3 \(8\pi r\)
4 \(16\pi r\)
Explanation:
According to de Broglie’s quantisation condition, \(2\pi r = n\lambda \) \( \Rightarrow \lambda = \frac{{2\pi r}}{n}\) \(r\,\alpha \,{n^2}\) \(\frac{{{r_1}}}{{{r_2}}} = {\left( {\frac{1}{4}} \right)^2} \Rightarrow {r_2} = 16{r_1}\) Hence, \(\lambda = \frac{{2\pi [16{r_1}]}}{4} = 8\pi {r_1} = 8\pi r\)
PHXII12:ATOMS
356524
Assertion : Smoky flame of Bunsen burner gives continuous spectrum whereas its blue flame gives band spectrum. Reason : The band spectrum consists of coloured bands of light on a dark background.
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.
Explanation:
When the flame of a Bunsen burner appears smoky, it indicates that there are carbon particles within it that are in an incandescent state. As a result, the flame emits (like grey body)a continuous spectrum of light. However, when the burner produces a blue flame, it signifies the presence of carbon and cyanogen, in a molecular state. This type of flame yields a band spectrum of light. So correct option is (2).
