Question

A photon of wavelength 1240 Å falls on photo sensitive material with work function of 2eV. The photoelectron either escape the material or collide with one of the atom of material and looses 10% of energy. The electron can make maximum n collision and can still escape the material then n is- [Given : $log_{0.9}(0.2) = 15.27$]

• A. 10
• B. 15
• C. 16
• D. 31

Answer 1

Answer: (A)

10

Answer 2

1. Photon Energy Calculation: The energy of the incident photon is given by $E = \frac{hc}{\lambda}$. Using the formula $E = \frac{12400}{\lambda}$ (where $E$ is in eV and $\lambda$ is in Å): $E_{photon} = \frac{12400 \text{ eV Å}}{1240 \text{ Å}} = 10 \text{ eV}$

2. Maximum Kinetic Energy of Photoelectron: According to Einstein's photoelectric equation, $KE_{max} = E_{photon} - \phi$. Given $\phi = 2 \text{ eV}$: $KE_{max} = 10 \text{ eV} - 2 \text{ eV} = 8 \text{ eV}$

3. Energy Loss in Collisions: After each collision, the electron retains 90% of its energy. After $k$ collisions, the kinetic energy $KE_k$ is: $KE_k = KE_{max} \times (0.9)^k$

4. Condition for Electron Escape: For the electron to escape, its kinetic energy must be at least equal to the work function ($\phi$). The problem states that the electron can make a maximum of $n$ collisions and still escape. This means: $KE_n \ge \phi$ $KE_{max} \times (0.9)^n \ge \phi$ Substituting the values: $8 \times (0.9)^n \ge 2$ $(0.9)^n \ge \frac{2}{8}$ $(0.9)^n \ge 0.25$

5. Determining the Maximum 'n' from Options: We need to find the largest integer $n$ from the given options that satisfies $(0.9)^n \ge 0.25$. Let's check each option:

   • A. n = 10: $(0.9)^{10} \approx 0.3487$. Since $0.3487 \ge 0.25$, this is a possible value for $n$.
   • B. n = 15: $(0.9)^{15} \approx 0.2059$. Since $0.2059 < 0.25$, this is not a possible value for $n$.
   • C. n = 16: $(0.9)^{16} \approx 0.1853$. Since $0.1853 < 0.25$, this is not a possible value for $n$.
   • D. n = 31: $(0.9)^{31} \approx 0.0386$. Since $0.0386 < 0.25$, this is not a possible value for $n$.

   Only $n=10$ satisfies the condition. Therefore, it is the maximum possible value among the given options for which the electron can still escape.

   (Note: Mathematically, solving $(0.9)^n \ge 0.25$ yields $n \le log_{0.9}(0.25)$. Using the given $log_{0.9}(0.2) = 15.27$ and change of base, $log_{0.9}(0.25) \approx 13.15$. This implies the maximum integer $n$ is 13. However, 13 is not an option. Given the options, $n=10$ is the only one that satisfies the condition.)