Q. No.Line corresponding to lyman series are \(L_1,L_2,L_3,L_4.....,\) among these \(L_1\) line corresponds to lowest energy. Similarly lines corresponding to balmer series are \(B_1,B_2,B_3,B_4....,\) among these \(B_1\) line corresponds to lowest energy:
\(\Delta E_L\) = Energy of \(1^{st}\) line of lyman series
\(\Delta E_B\) = Energy of \(1^{st}\) line of balmer series
If \(\Delta E_L=x.\Delta E_B\). Calculate \((x\times 10^{-1})\):
1. 54
2. 27
3. 40
4. 18
\(\Delta E_L\) = Energy of \(1^{st}\) line of lyman series
\(\Delta E_B\) = Energy of \(1^{st}\) line of balmer series
If \(\Delta E_L=x.\Delta E_B\). Calculate \((x\times 10^{-1})\):
1. 54
2. 27
3. 40
4. 18
Subtopic: Electromagnetic Radiation | Type of Spectra |
64%
Level 2: 60%+
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Heat of atomisation of \(CH_4(g) \) and \(C_2H_6(g)\) are x kJ/mole and y kJ/mole. Find the maximum wavelength of photon required to dissociate (C–C) bond in \(C_2H_6\):
1. \(\frac{\mathrm{hcN}_{\mathrm{A}}}{\left[\mathrm{y}-\frac{3 \mathrm{x}}{2}\right]} \)
2. \(\frac{\mathrm{hcN}_{\mathrm{A}}}{\left[\frac{4 \mathrm{x}-6 \mathrm{y}}{4}\right]} \)
3. \(\frac{\mathrm{hcN}_{\mathrm{A}}}{250\left[\frac{3 \mathrm{x}}{2}-\mathrm{y}\right]} \)
4. \(\frac{\mathrm{hcN}_{\mathrm{A}}}{500[2 \mathrm{y}-3 \mathrm{x}]}\)
1. \(\frac{\mathrm{hcN}_{\mathrm{A}}}{\left[\mathrm{y}-\frac{3 \mathrm{x}}{2}\right]} \)
2. \(\frac{\mathrm{hcN}_{\mathrm{A}}}{\left[\frac{4 \mathrm{x}-6 \mathrm{y}}{4}\right]} \)
3. \(\frac{\mathrm{hcN}_{\mathrm{A}}}{250\left[\frac{3 \mathrm{x}}{2}-\mathrm{y}\right]} \)
4. \(\frac{\mathrm{hcN}_{\mathrm{A}}}{500[2 \mathrm{y}-3 \mathrm{x}]}\)
Subtopic: Electromagnetic Radiation |
Level 4: Below 35%
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Energy of \(1^{st}\) Balmer line of H-atom is x J. The energy (in J) of second Balmer line of H-atom is :
1. x
2. 2x
3. 1.35x
4. \(\dfrac{\mathrm{x}}{1.35}\)
1. x
2. 2x
3. 1.35x
4. \(\dfrac{\mathrm{x}}{1.35}\)
Subtopic: Electromagnetic Radiation | Type of Spectra |
75%
Level 2: 60%+
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On two metal surfaces, a monochromatic light of 6 eV was incident. They have ratio of their work function and maximum KE as
\(\frac{\phi_1}{\phi_2}=\frac{1}{2}, \frac{\left(\mathrm{KE}_{\max }\right)_1}{\left(\mathrm{KE}_{\max }\right)_2}=\frac{2.62}{1}\)
Then \(\phi_1\) and \(\phi_2\) values are respectively (in eV):
1. 2.292, 4.584
2. 4.584, 2.292
3. 4.584, 9.168
4. 1.146, 2.292
\(\frac{\phi_1}{\phi_2}=\frac{1}{2}, \frac{\left(\mathrm{KE}_{\max }\right)_1}{\left(\mathrm{KE}_{\max }\right)_2}=\frac{2.62}{1}\)
Then \(\phi_1\) and \(\phi_2\) values are respectively (in eV):
1. 2.292, 4.584
2. 4.584, 2.292
3. 4.584, 9.168
4. 1.146, 2.292
Subtopic: Photo Electric Effect | Electromagnetic Radiation |
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Level 3: 35%-60%
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Find out wavelength of a photon having frequency equal to \(900 ~\text {sec}^{-1}\)
1. \(3.33 \times 10^5 \mathrm{~m} \)
2. \(3.33 \times 10^5 \mathrm{~cm} \)
3. \(3.33 \times 10^7 \mathrm{~m} \)
4. \(3.33 \times 10^4 \mathrm{~m}\)
1. \(3.33 \times 10^5 \mathrm{~m} \)
2. \(3.33 \times 10^5 \mathrm{~cm} \)
3. \(3.33 \times 10^7 \mathrm{~m} \)
4. \(3.33 \times 10^4 \mathrm{~m}\)
Subtopic: Electromagnetic Radiation |
79%
Level 2: 60%+
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Frequency of following electromagnetic wave is given by ______ x \(10^6 \mathrm{~Hz} .\)
1. 25
2. 67
3. 98
4. 23
1. 25
2. 67
3. 98
4. 23
Subtopic: Electromagnetic Radiation |
86%
Level 1: 80%+
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Wave number of a radiation having 5800Å wavelength is x\(\mathrm{x} \times 10^4 \mathrm{~cm}^{-1} \text {. }\) The value of \(\mathrm{x}\) to nearest integer is :
1. 2
2. 4
3. 5
4. 8
1. 2
2. 4
3. 5
4. 8
Subtopic: Electromagnetic Radiation |
83%
Level 1: 80%+
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If the energy of radiation having wavelength of 242 nm is \(x\times10^{-19}\text{J},\) then find nearest integer value of x.
Given : Plank's constant = \(6.6\times10^{-19}\) Joule-Second, \(\text{c}=3\times10^{8}\text{ms}^{-1}\)
1. 8
2. 2
3. 5
4. 2
Given : Plank's constant = \(6.6\times10^{-19}\) Joule-Second, \(\text{c}=3\times10^{8}\text{ms}^{-1}\)
1. 8
2. 2
3. 5
4. 2
Subtopic: Electromagnetic Radiation |
83%
Level 1: 80%+
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The energy of radiation given by \(E=\frac{hc}{\lambda_{\text{absorb}}}. \)
If \(E=+96\text{ kJ}mol^{-1}\), then the value of \(\lambda_{\text{absorb}} \) (in \(A^\circ\)) is:
1. 12471 \(A^\circ\)
2. 124.71 \(A^\circ\)
3. 1247.1 \(A^\circ\)
4. 1.2471 \(A^\circ\)
If \(E=+96\text{ kJ}mol^{-1}\), then the value of \(\lambda_{\text{absorb}} \) (in \(A^\circ\)) is:
1. 12471 \(A^\circ\)
2. 124.71 \(A^\circ\)
3. 1247.1 \(A^\circ\)
4. 1.2471 \(A^\circ\)
Subtopic: Electromagnetic Radiation |
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Level 2: 60%+
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The total energy of 1 mol of photons in J/mol having \(\lambda = 600 \ nm \) is:
Given: \(h = 6.62 \times 10^{-34} \mathrm{~J} \ \mathrm{sec}, \mathrm{c}=3 \times 10^{8} \mathrm{~m \ s}^{-1}\)
1. \(2 \times 10^5 J/mol\)
2. \(6.64 \times 10^8 J/mol\)
3. \(1.24 \times 10^4 J/mol\)
4. \(1.24 \times 10^8 J/mol\)
Given: \(h = 6.62 \times 10^{-34} \mathrm{~J} \ \mathrm{sec}, \mathrm{c}=3 \times 10^{8} \mathrm{~m \ s}^{-1}\)
1. \(2 \times 10^5 J/mol\)
2. \(6.64 \times 10^8 J/mol\)
3. \(1.24 \times 10^4 J/mol\)
4. \(1.24 \times 10^8 J/mol\)
Subtopic: Electromagnetic Radiation |
69%
Level 2: 60%+
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