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:
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
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Subtopic: Electromagnetic Radiation |
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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
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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
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Subtopic: Electromagnetic Radiation | Type of Spectra |
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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
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Subtopic: Photo Electric Effect | Electromagnetic Radiation |
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