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Two radiations of photons energies 1 eV and 2.5 eV, successively illuminate a photosensitive metallic surface of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons is

A

`1:4`

B

`1:2`

C

`1:1`

D

`1:5`

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The correct Answer is:
To solve the problem, we need to determine the ratio of the maximum speeds of the emitted electrons when two different photon energies illuminate a photosensitive metallic surface. The work function of the surface is given, and we will use the photoelectric effect principles to find the solution. ### Step-by-Step Solution: 1. **Identify the Given Values:** - Photon energy 1 (E1) = 1 eV - Photon energy 2 (E2) = 2.5 eV - Work function (Φ) = 0.5 eV 2. **Calculate the Kinetic Energy of Electrons:** The kinetic energy (KE) of the emitted electrons can be calculated using the formula: \[ KE = E - \Phi \] where \(E\) is the energy of the incident photon. - For the first photon (E1 = 1 eV): \[ KE_1 = E1 - \Phi = 1 \, \text{eV} - 0.5 \, \text{eV} = 0.5 \, \text{eV} \] - For the second photon (E2 = 2.5 eV): \[ KE_2 = E2 - \Phi = 2.5 \, \text{eV} - 0.5 \, \text{eV} = 2.0 \, \text{eV} \] 3. **Relate Kinetic Energy to Speed:** The kinetic energy of an electron is also given by the formula: \[ KE = \frac{1}{2} mv^2 \] where \(m\) is the mass of the electron and \(v\) is its speed. 4. **Set Up the Ratio of Kinetic Energies:** We can set up the ratio of the kinetic energies: \[ \frac{KE_1}{KE_2} = \frac{0.5 \, \text{eV}}{2.0 \, \text{eV}} = \frac{1}{4} \] 5. **Relate the Speeds:** Using the relationship between kinetic energy and speed, we can write: \[ \frac{KE_1}{KE_2} = \frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} = \frac{v_1^2}{v_2^2} \] Since the mass \(m\) and the factor \(\frac{1}{2}\) cancel out. 6. **Calculate the Ratio of Speeds:** From the ratio of kinetic energies: \[ \frac{v_1^2}{v_2^2} = \frac{1}{4} \] Taking the square root of both sides gives: \[ \frac{v_1}{v_2} = \frac{1}{2} \] ### Final Result: The ratio of the maximum speeds of the emitted electrons is: \[ \frac{v_1}{v_2} = \frac{1}{2} \]

To solve the problem, we need to determine the ratio of the maximum speeds of the emitted electrons when two different photon energies illuminate a photosensitive metallic surface. The work function of the surface is given, and we will use the photoelectric effect principles to find the solution. ### Step-by-Step Solution: 1. **Identify the Given Values:** - Photon energy 1 (E1) = 1 eV - Photon energy 2 (E2) = 2.5 eV - Work function (Φ) = 0.5 eV ...
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