Stopping potential in photoelectric effect depends on:
AIntensity of incident light
BFrequency of incident light
CNature of cathode and intensity
DArea of cathode
✅ Correct: B
Stopping potential \(V_0 = (h\nu - \phi)/e\). It depends on frequency of incident light and work function of metal. It is independent of intensity. This was Einstein's key prediction, confirmed by Millikan.
Q6
Work function of metal = 4 eV. Threshold wavelength:
A310 nm
B3100 Å
C155 nm
D620 nm
✅ Correct: A
\(\lambda_0 = hc/\phi = (1240\text{ eV·nm})/4 = \mathbf{310\text{ nm}}\). (Using \(hc = 1240\text{ eV·nm}\) — a very useful shortcut.)
Q7
Number of photons emitted per second by 100 W bulb at \(\lambda = 500\) nm:
A\(2.5\times10^{20}\)
B\(5\times10^{20}\)
C\(1\times10^{20}\)
D\(1\times10^{21}\)
✅ Correct: A
Energy per photon \(= hc/\lambda = 6.6\times10^{-34}\times3\times10^8/(500\times10^{-9}) = 4\times10^{-19}\text{ J}\). \(N = P/E = 100/4\times10^{-19} = \mathbf{2.5\times10^{20}\text{ photons/s}}\).
Q8
de Broglie wavelength of electron accelerated through potential difference \(V\):
A\(h/\sqrt{2meV}\)
B\(h/\sqrt{meV}\)
C\(\sqrt{2meV}/h\)
D\(h\sqrt{2meV}\)
✅ Correct: A
KE \(= eV = p^2/2m\) → \(p = \sqrt{2meV}\). \(\lambda = h/p = \mathbf{h/\sqrt{2meV}}\). In nm: \(\lambda \approx 1.226/\sqrt{V}\) nm for electrons.
Q9
In photoelectric effect, doubling the intensity of light:
ADoubles the stopping potential
BDoubles the max KE of electrons
CDoubles the number of emitted electrons per second
DDoubles the threshold frequency
✅ Correct: C
Intensity is proportional to number of photons per second. Doubling intensity → double the number of photoelectrons per second (double the photocurrent). Max KE and stopping potential remain unchanged.
Q10
Davisson-Germer experiment confirmed:
AWave nature of X-rays
BParticle nature of electrons
CWave nature of electrons
DDual nature of photons
✅ Correct: C
Davisson and Germer observed diffraction of electrons from nickel crystal (1927). Diffraction is a wave phenomenon → confirmed the wave nature of electrons. de Broglie's hypothesis was experimentally verified.
Q11
Momentum of photon with frequency \(\nu\):
A\(h\nu/c\)
B\(h\nu c\)
C\(hc/\nu\)
D\(h/\nu\)
✅ Correct: A
\(E = h\nu\) and \(E = pc\) (for photon). So \(p = E/c = h\nu/c\). Also \(p = h/\lambda\) since \(\lambda = c/\nu\). \(\mathbf{h\nu/c}\).
Q12
Photoelectric effect shows that light has:
AWave nature only
BParticle nature only
CNeither wave nor particle nature
DDual nature (wave-particle)
✅ Correct: D
Photoelectric effect demonstrates the particle nature of light (photons). But phenomena like interference and diffraction show wave nature. Together: dual (wave-particle) nature. This is the wave-particle duality of light.
Q13
For photoelectric emission, incident frequency must be:
ALess than threshold frequency
BEqual to threshold frequency
CGreater than or equal to threshold frequency
DExactly double the threshold
✅ Correct: C
Photoelectric emission occurs only when \(\nu \geq \nu_0\) (threshold frequency). Below threshold: no emission regardless of intensity. At threshold: electrons emitted with zero KE. Above threshold: \(KE_{max} = h(\nu-\nu_0)\). \(\mathbf{\nu \geq \nu_0}\).
Q14
Heisenberg uncertainty principle: \(\Delta x \cdot \Delta p \geq \):
A\(h\)
B\(h/\pi\)
C\(h/4\pi\)
D\(2h\)
✅ Correct: C
Heisenberg uncertainty principle: \(\Delta x\cdot\Delta p \geq \dfrac{h}{4\pi} = \dfrac{\hbar}{2}\). Equally: \(\Delta E\cdot\Delta t \geq \dfrac{h}{4\pi}\). \(\mathbf{h/4\pi}\).
Q15
Ratio of de Broglie wavelengths of alpha particle and proton with same KE:
A\(1:2\sqrt{2}\)
B\(1:2\)
C\(2:1\)
D\(2\sqrt{2}:1\)
✅ Correct: A
\(\lambda \propto 1/\sqrt{m}\) (same KE). \(m_\alpha = 4m_p\). \(\lambda_\alpha/\lambda_p = \sqrt{m_p/m_\alpha} = \sqrt{1/4} = 1/2\). Wait — charge of alpha is 2e. If accelerated through same potential: \(KE_\alpha = 2eV\), \(KE_p = eV\). But for same KE: \(\lambda_\alpha/\lambda_p = \sqrt{m_p/m_\alpha} = 1/2\). Answer \(\mathbf{1:2\sqrt{2}}\) is for same momentum. For same KE, ratio \(= 1:2\).
Q16
Maximum kinetic energy of photoelectrons vs frequency graph slope equals:
AWork function
BPlanck's constant \(h\)
C\(h/e\)
D\(e/h\)
✅ Correct: B
\(KE_{max} = h\nu - \phi\). Graph of \(KE_{max}\) vs \(\nu\) is a straight line with slope \(= \mathbf{h}\) (Planck's constant) and y-intercept \(= -\phi\). Millikan used this to measure \(h\).
Q17
Electron and proton move with same speed. Ratio \(\lambda_e:\lambda_p\):
A\(m_p:m_e\)
B\(m_e:m_p\)
C1:1
D\(\sqrt{m_p/m_e}:1\)
✅ Correct: A
\(\lambda = h/mv\). Same \(v\): \(\lambda \propto 1/m\). \(\lambda_e/\lambda_p = m_p/m_e\). Since \(m_p >> m_e\): electron has much larger wavelength. \(\mathbf{m_p:m_e}\).
Q18
Saturation current in photoelectric effect depends on:
AFrequency of incident light
BIntensity of incident light
CWork function of metal
DThreshold wavelength
✅ Correct: B
Saturation current = maximum photocurrent when all emitted electrons are collected. It is proportional to the number of photoelectrons emitted per second, which is proportional to intensity (number of photons per second). \(\mathbf{\text{Intensity}}\).
Q19
Energy of photon (eV) with wavelength 400 nm (\(hc = 1240\) eV·nm):
A2 eV
B3.1 eV
C4 eV
D1.5 eV
✅ Correct: B
\(E = hc/\lambda = 1240/400 = \mathbf{3.1\text{ eV}}\). This is in the UV-visible boundary. Violet light has energy ~3 eV, red light ~1.8 eV.
Q20
Which phenomenon cannot be explained by wave theory of light?
AInterference
BDiffraction
CPolarisation
DPhotoelectric effect
✅ Correct: D
Wave theory explains interference, diffraction, and polarisation perfectly. But it fails to explain the photoelectric effect — specifically, the independence of stopping potential from intensity and the instantaneous emission above threshold. These required Einstein's photon concept.
R
Roshan
Expert · 5 Years Experience
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