Q.The photoelectric cut-off voltage in a certain experiment is $1.5\ \text{V}$. What is the maximum kinetic energy of photoelectrons emitted?

The key concept is the photoelectric effect and the definition of stopping potential. When light shines on a metal, electrons are ejected with a range of kinetic energies, up to a maximum. The stopping potential is the voltage that just barely stops the fastest electrons — it directly measures their maximum kinetic energy in electron-volts. No work function or photon energy calculation is needed here because the maximum kinetic energy is already given.

1. Recall the photoelectric equation:

   The maximum kinetic energy of emitted electrons is $K_{\text{max}} = hf - \phi$, where $\phi$ is the work function. But the stopping potential $V_s$ is defined by $e V_s = K_{\text{max}}$. That is, the stopping potential (in volts) numerically equals the maximum kinetic energy (in electron-volts).

2. Apply the given data:

   The problem states that the maximum kinetic energy is $0.4 \, \text{eV}$. Therefore,

3. Ignore the work function: The work function (1.4 eV) is extra information — it would be needed to find the photon energy, but not for the stopping potential once $K_{\text{max}}$ is known.

Concept & Intuition

The photoelectric effect tells us that a single photon gives all its energy $h\nu$ to a single electron. The electron uses part of that energy to overcome the work function $\phi$ (the minimum energy needed to escape), and the remainder becomes kinetic energy. The maximum kinetic energy occurs for electrons that are ejected from the very surface (no energy lost to collisions inside the material). That maximum kinetic energy is simply

The intensity of the beam tells us how many photons arrive per second per square meter, and the “0.53% efficiency” tells us what fraction of those photons actually eject electrons — but neither affects the maximum speed of any individual electron. The speed depends only on the energy difference $h\nu - \phi$. So we ignore the intensity and efficiency for this part of the problem.

Step-by-step solution

1. Convert energies to joules Photon energy: $E_{\text{photon}} = 11.2 \, \text{eV}$. Work function: $\phi = 4.8 \, \text{eV}$. Use $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$.

2. Find maximum kinetic energy

3. Relate kinetic energy to speed For a non‑relativistic electron (speed much less than $c$),

Given $m = 9 \times 10^{-31} \, \text{kg}$,

4. Calculate First compute numerator: $2 \times 1.024 \times 10^{-18} = 2.048 \times 10^{-18}$. Divide by mass:

Take square root:

Rounding to two significant figures gives $1.5 \times 10^6 \, \text{m/s}$.

The key concept here is the photoelectric effect equation:

where $K_{\text{max}}$ is the maximum kinetic energy of emitted photoelectrons, $h$ is Planck’s constant, $f$ is the frequency of incident radiation, and $\phi$ is the work function of the metal (the minimum energy needed to eject an electron).

When we plot $K_{\text{max}}$ on the y-axis and $f$ on the x-axis, the equation is of the form $y = mx + c$:

Here:

• Slope $m = h$ (a universal constant)
• y-intercept $c = -\phi$

The intercept on the $K_{\text{max}}$-axis (when $f = 0$) is $-\phi$. But the question asks about the intercept on the frequency axis — that is, the value of $f$ when $K_{\text{max}} = 0$. Let’s clarify.

1. Identify which intercept is meant The phrase “intercept of which” in the problem refers to the point where the straight line meets the frequency axis (the x-axis). At that point, $K_{\text{max}} = 0$. Set $K_{\text{max}} = 0$ in the equation:

This $f_0$ is the threshold frequency — the minimum frequency needed to eject electrons.

2. What does the threshold frequency depend on?

   • $\phi$ is the work function, which is a property of the metal (different metals have different work functions).
   • $h$ is Planck’s constant, the same for all materials. Therefore, $f_0 = \phi / h$ depends only on the nature of the metal.

3. Does intensity affect the intercept?

   Intensity of radiation changes the number of photons per second, not the energy of individual photons. It affects the number of emitted electrons (photocurrent), but not the maximum kinetic energy or the threshold frequency. So intensity has no effect on the intercept.

4. Classic pitfall

   Watch out

   A common mistake is to think the intercept is the work function itself. But the intercept on the frequency axis is the threshold frequency, which is proportional to the work function. Both depend only on the metal.

5. Conclusion

   The intercept (threshold frequency) depends on the nature of the metal used, not on intensity. That matches option A.

Step 1 — Einstein's photoelectric equation

A single photon of energy $E$ is absorbed by a single electron. Part of that energy pays the work function $\phi_0$ (the minimum energy needed to just liberate an electron from the metal surface); whatever is left appears as kinetic energy:

Step 2 — Substitute the given work function

Also note $E > \phi_0$, so emission does occur — the incident light is above threshold.

Step 3 — The conceptual point: why $K_{\max}$ and not $K$

The subscript "max" is not decoration. The work function $\phi_0$ is the minimum energy required — that of the least tightly bound electrons, those right at the surface.

• An electron sitting at the surface, needing exactly $\phi_0$ and suffering no collision on the way out, emerges with the full $K_{\max}=2E/3$.
• An electron from deeper inside the metal is more tightly bound and/or loses energy in collisions with the lattice and other electrons on its way to the surface. It emerges with less than $K_{\max}$ — possibly with essentially zero KE.

So the photoelectrons are emitted with a continuous distribution of kinetic energies:

Step 4 — Eliminate the options

• (A) $K = \tfrac{2E}{3}$ — this is the correct maximum, but wrongly asserts that every photoelectron has exactly this energy. That is the standard trap.
• (B) $K = \tfrac{E}{3}$ — this is $\phi_0$ itself, not the kinetic energy.
• (D) $0\le K\le \tfrac{E}{3}$ — the right form (a range) but the wrong ceiling; the ceiling is $E-\phi_0$, not $\phi_0$.
• (C) correctly gives both the range and the ceiling $2E/3$.

ANSWER: C

Einstein's photoelectric equation: $K=\dfrac{hc}{\lambda}-W$.

With $\lambda_1=3\lambda_2$:

Compute $\tfrac13 K_2=\dfrac{hc}{3\lambda_2}-\dfrac{W}{3}$. Subtract:

Therefore $K_1<\dfrac{1}{3}K_2$.

Einstein's photoelectric equation: $K=\dfrac{hc}{\lambda}-\phi$.

Compare $2K_1$ with $K_2$:

Thus $2K_1<K_2$, i.e. $K_1<\dfrac{K_2}{2}$.

Concept: Einstein's photoelectric equation, K = hc/lambda - phi (same metal, so the same work function phi > 0).

K1 = hc/lambda1 - phi

K2 = hc/lambda2 - phi

Given lambda1 = 3 lambda2, i.e. lambda2 = lambda1/3, so hc/lambda2 = 3 hc/lambda1.

K2 = 3 (hc/lambda1) - phi = 3 (K1 + phi) - phi = 3 K1 + 2 phi

Since phi > 0:

K2 > 3 K1

=> K1 < K2 / 3

(Options C and D cannot be asserted in general - they would need a specific value of phi.)

ANSWER: B

Step 1 — The detection condition

A photodiode works by photon absorption creating an electron–hole pair: the photon must lift an electron from the valence band across the forbidden gap into the conduction band. That is only possible if

A photon with less energy than the gap simply cannot make the transition — the material is transparent to it and no photocurrent flows, however intense the light.

Step 2 — Photon energy of 600 nm light

Use $E = \dfrac{hc}{\lambda}$, with the convenient form $E(\text{eV}) = \dfrac{1240}{\lambda(\text{nm})}$ (since $hc \approx 1240\ \text{eV·nm}$):

(Cross-check in SI: $E = \frac{6.6\times10^{-34}\times 3\times10^{8}}{600\times10^{-9}} = 3.3\times10^{-19}\ \text{J} = \frac{3.3\times10^{-19}}{1.6\times10^{-19}} = 2.06\ \text{eV}$ ✓)

Step 3 — Compare with each diode's band gap

Only $D_2$ clears its band gap — and only just (by $0.07$ eV), which is exactly the kind of margin the question is testing.

Step 4 — The physical takeaway

Each photodiode has a maximum detectable wavelength (a cut-off), $\lambda_{max} = \dfrac{1240}{E_g(\text{eV})}$ nm:

• $D_1$: $\lambda_{max} = 1240/2.5 = 496$ nm
• $D_2$: $\lambda_{max} = 1240/2 = 620$ nm
• $D_3$: $\lambda_{max} = 1240/3 = 413$ nm

Our $600$ nm light lies below $620$ nm only for $D_2$. Same conclusion, viewed from the wavelength side. ✓

ANSWER: C

By Einstein's photoelectric equation, the maximum kinetic energy is

Case 1: $E = 2\phi \Rightarrow KE_1 = 2\phi - \phi = \phi$.

Case 2: $E = 5\phi \Rightarrow KE_2 = 5\phi - \phi = 4\phi$.

Since $KE = \tfrac12 m v^2$, $v \propto \sqrt{KE}$:

The key idea here is that light exerts a force because photons carry momentum. When a photon is absorbed by a surface, its momentum is transferred to that surface. If the surface is "completely darkened," it means every photon that hits it is absorbed — none are reflected. So the force you measure is simply the rate at which momentum is delivered by the incoming stream of photons.

The momentum of a single photon is given by de Broglie's relation:

where $h$ is Planck's constant and $\lambda$ is the wavelength.

If $n$ photons fall per second, the total momentum transferred per second is $n \cdot p$. And force is exactly the rate of change of momentum:

Now we just plug in the numbers.

1. Write the force equation

2. Substitute the given values $F = 6.62 \times 10^{-5}$ N, $h = 6.62 \times 10^{-34}$ J·s, $\lambda = 5 \times 10^{-7}$ m

3. Simplify the fraction

4. Solve for $n$

The question asks for $n$ in units of $10^{22}$, so $n = 5 \times 10^{22}$ means the blank is filled with 5.

Step 1 — Einstein's photoelectric equation.

Light of frequency $\nu$ arrives as photons each of energy $E=h\nu$. One photon is absorbed by one electron. Part of that energy ($\phi_0$, the work function) is spent escaping the metal surface; the rest appears as kinetic energy. For the most loosely bound electrons:

So $K_{\max}$ is a straight-line function of $\nu$ with slope $h$ — increase the frequency and each electron leaves faster.

Step 2 — Why not intensity / number of photons (options A and D)?

Intensity = number of photons per second. Doubling it doubles the number of electrons emitted (the photocurrent) but each photon still carries the same $h\nu$, so each electron still leaves with the same $K_{\max}$. This is precisely the observation classical wave theory could not explain — and it is why the stopping potential is independent of intensity.

Step 3 — Why not speed (option C)?

The speed of the incident radiation in vacuum is $c$ for all light, so it cannot distinguish one beam from another; it cannot control $K_{\max}$.

Step 4 — Commit.

Only frequency (through $h\nu$) appears in the expression for $K_{\max}$. Below the threshold frequency $\nu_0=\phi_0/h$ no electrons are emitted at all, however intense the beam.

ANSWER: B