A convex lens of focal length 40 cm$40\text{ cm}$ forms an image of an extended source of light on a photoelectric cell. A current I$I$ is produced. The lens is replaced by another convex lens having the same diameter but focal length 20 cm$20\text{ cm}$. The photoelectric current now is:

Related Formula

For a photon, the energy (which is its kinetic energy) is:

For an electron, the de-Broglie wavelength related to its kinetic energy (Kₑ$K_e$) is:

since Kₑ = (1)/(2) mₑ vₑ² mₑ vₑ = (2 Kₑ)/(vₑ)$K_e = \frac{1}{2} m_e v_e^2 \implies m_e v_e = \frac{2 K_e}{v_e}$.

Core Logic

Given that the velocity of the electron is 25%$25\%$ of the speed of light:

Substituting this into the electron's wavelength equation:

Step 1: Equate Wavelengths

We are given that the de-Broglie wavelengths are equal (λₚ = λₑ$\lambda_p = \lambda_e$):

Cancelling h c$h c$ from both sides:

Step 2: Find the Energy Ratio

The ratio of the kinetic energy of the electron to that of the photon (Eₚ$E_p$) is:

Therefore, the ratio is 1/8$1/8$.

Pattern Recognition

A direct comparison between massive particle kinetic energy (K = (p²)/(2m) = (p v)/(2)$K = \frac{p^2}{2m} = \frac{p v}{2}$) and photon energy (E = p c$E = p c$) when they share the same momentum p$p$ (same wavelength) yields a simple universally applicable rule: KparticleEphoton = (v)/(2c)$\frac{K_{\text{particle}}}{E_{\text{photon}}} = \frac{v}{2c}$.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Related Formula

Core Logic

From Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by Ek = hν - φ$E_k = h\nu - \phi$, where h$h$ is Planck's constant, ν$\nu$ is the incident frequency, and φ$\phi$ is the work function. This represents a straight line equation of the form y = mx + c$y = mx + c$, where y = Ek$y = E_k$ and x = ν$x = \nu$.

Step 1: Identify the Slope

Comparing Ek = hν - φ$E_k = h\nu - \phi$ with y = mx + c$y = mx + c$, we get the slope:

Thus, the slope of the graph gives Planck's constant.

Pattern Recognition

If the y-axis is K.E.$K.E.$, the slope is h$h$. If the y-axis is Stopping Potential V₀$V_0$, the equation is eV₀ = hν - φ V₀ = (h/e)ν - (φ/e)$eV_0 = h\nu - \phi \implies V_0 = (h/e)\nu - (\phi/e)$, and the slope would be h/e$h/e$.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Related Formula

Core Logic

For photoelectric emission to occur, the incident energy must be greater than or equal to the work function (Wₑ$W_e$ or Φ$\Phi$). The longest wavelength corresponds to the minimum energy required, which is exactly the work function.

Step 1: Calculate Wavelength

Substitute the given values using the convenient constant hc ≈ 1240 ~eV · nm$hc \approx 1240 \mathrm{~eV \cdot nm}$:

The maximum wavelength (longest wavelength) is:

Pattern Recognition

The standard shortcut E(eV) = 1240 / λ(nm)$E(\text{eV}) = 1240 / \lambda(\text{nm})$ rapidly converts work function to threshold wavelength without dealing with standard SI unit conversions.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Related Formula

For photoelectric emission to occur: f ≥ f₀$f \ge f_0$ where f$f$ is incident frequency and f₀$f_0$ is the threshold frequency.

Core Logic

Initially, the frequency is f₁ = 1.5 f₀$f_1 = 1.5 f_0$. Since f₁ > f₀$f_1 > f_0$, emission happens. Then, the frequency is halved: f₂ = (1.5 f₀)/(2) = 0.75 f₀$f_2 = \frac{1.5 f_0}{2} = 0.75 f_0$.

Step 1: Check Threshold Condition

Since f₂ = 0.75 f₀$f_2 = 0.75 f_0$, we see that f₂ < f₀$f_2 < f_0$. The incident light no longer has enough energy per photon to overcome the work function, regardless of how intense the light is.

Step 2: Conclusion

Because the threshold condition fails, emission completely stops. The number of photoelectrons emitted is zero.

Pattern Recognition

Always check the frequency threshold first in photoelectric questions. Intensity adjustments are irrelevant traps if f < f₀$f < f_0$. No emission occurs below threshold frequency.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter