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Q. No.Select the correct option based on the statements given below:
Statement I:
By de-Broglie's hypothesis momentum of an electron, \(p=h/ \lambda\).
Statement II:
The energy of an electron is given by; \(E=hc/ \lambda\).
1.
Statement I is correct and Statement II is incorrect.
2.
Statement I is incorrect and Statement II is correct.
3.
Both Statement I and Statement II are correct.
4.
Both Statement I and Statement II are incorrect.
Subtopic: De-broglie Wavelength |
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The ratio of wavelengths of proton and deuteron accelerated by potential \(V_{p}\) and \(V_{d}\) is \(1:\sqrt2.\) Then, the ratio of \(V_{p}\) to \(V_{d}\) will be:
1. \(1:1\)
2. \(\sqrt 2: 1\)
3. \(2:1\)
4. \(4:1\)
1. \(1:1\)
2. \(\sqrt 2: 1\)
3. \(2:1\)
4. \(4:1\)
Subtopic: De-broglie Wavelength |
From NCERT
JEE
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A nucleus of mass \(M,\) initially at rest, splits into two fragments with masses \(\dfrac{M'}{ 3}\) and \(\dfrac{2M'} { 3}\) \((M'<M).\) The ratio of the de-Broglie wavelengths of the two fragments is:
| 1. | \(1:2\) | 2. | \(2:1\) |
| 3. | \(1:1\) | 4. | \(2:3\) |
Subtopic: De-broglie Wavelength |
56%
From NCERT
JEE
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Let \(K_{1}\) and \(K_{2}\) be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength \(\lambda_1\) and \(\lambda_2\), respectively, are incident on a metallic surface. If \(\lambda_1 = 3 \lambda_2,\) then:
| 1. | \(K_1>\dfrac{K_2}{3} \) | 2. | \({K}_1<\dfrac{{K}_2}{3} \) |
| 3. | \({K}_1=\dfrac{{K}_2}{3} \) | 4. | \({K}_2=\dfrac{{K}_1}{3}\) |
Subtopic: Einstein's Photoelectric Equation |
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JEE
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The de-Broglie wavelength of a photon of energy \(E\) is \(\lambda_{ph}\) and that of an electron (non-relativistic) of the same energy \(E\) is \(\lambda_{e}.\) Then (assume \(E\text ~\)few \(e\text{V}\)):
1. \(\lambda_{ph}=\lambda_e\)
2. \(\lambda_{ph}<\lambda_e\)
3. \(\lambda_{ph}>\lambda_e\)
4. any of the above may be true
1. \(\lambda_{ph}=\lambda_e\)
2. \(\lambda_{ph}<\lambda_e\)
3. \(\lambda_{ph}>\lambda_e\)
4. any of the above may be true
Subtopic: De-broglie Wavelength |
From NCERT
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A photon of energy \(10.2\) eV corresponds to light of wavelength \(\lambda_0\). Due to electron transition from \(n = 2 \) to \(n = 1\) in a hydrogen atom, light of wavelength \(\lambda\) is emitted. If we take into account the recoil of atom when photon is emitted then:
| 1. | \(\lambda = \lambda_0\) |
| 2. | \(\lambda < \lambda_0\) |
| 3. | \(\lambda > \lambda_0\) |
| 4. | data is not sufficient to reach a conclusion |
Subtopic: Photoelectric Effect: Experiment |
From NCERT
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Hints
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\(A\) and \(B\) are two metals with threshold frequencies \(1.8\times 10^{14}\ \) Hz and \(2.2\times 10^{14}\ \) Hz. Two identical photons of energy \(0.825\) eV each are incident on them. Then, photoelectrons are emitted in:
(Take \(h=6.6\times 10^{-34}\ \) J-s)
1. \(B\) only
2. \(A\) only
3. neither \(A\) nor \(B\)
4. both \(A\) and \(B\)
(Take \(h=6.6\times 10^{-34}\ \) J-s)
1. \(B\) only
2. \(A\) only
3. neither \(A\) nor \(B\)
4. both \(A\) and \(B\)
Subtopic: Photoelectric Effect: Experiment |
From NCERT
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Photoelectrons emerging from a photocathode (work function: \(2.2~\text{eV}\)) are allowed to fall onto a gas containing hydrogen atoms in the ground state and the first excited state. What is the minimum energy of the photons incident on the photo-cathode that will cause the photoelectrons to transfer energy to the \(\mathrm{H\text-}\)atoms?
| 1. | \(13.6~\text{eV}+2.2~\text{eV}\) |
| 2. | \((10.2+2.2)~\text{eV}\) |
| 3. | \((3.4+2.2)~\text{eV}\) |
| 4. | \((1.89+2.2)~\text{eV}\) |
Subtopic: Photoelectric Effect: Experiment |
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The de-Broglie wavelength associated with an electron, accelerated by a potential difference of \(81\) V is given by:
1. \(13.6\) nm
2. \(136\) nm
3. \(1.36\) nm
4. \(0.136\) nm
1. \(13.6\) nm
2. \(136\) nm
3. \(1.36\) nm
4. \(0.136\) nm
Subtopic: De-broglie Wavelength |
58%
From NCERT
NEET - 2023
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A metallic ball (work function: \(2\) eV) is irradiated with light consisting of photons of wavelength \(200\) nm. The ball has an initial charge, giving it a potential \(1\) V. Take the product of Planck's constant and velocity of light, hc as \(1240\) eV-nm. The final potential of the ball, when photoemission practically stops, is:
1. \(2\) V
2. \(3.2\) V
3. \(4.2\) V
4. \(5.2\) V
1. \(2\) V
2. \(3.2\) V
3. \(4.2\) V
4. \(5.2\) V
Subtopic: Photoelectric Effect: Experiment |
54%
From NCERT
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