When light of sufficiently high frequency is incident on a metallic surface, electrons are emitted from the metallic surface. This phenomenon is called photoelectric emission. Kinetic energy of the emitted photoelectrons depends on the wavelength of incident light and is independent of the intensity of light. Number of emitted photoelectrons depends on intensity. `(hv-phi)` is the maximum kinetic energy of emitted photoelectron (where `phi` is the work function of metallic surface). Reverse effect of photo emission produces X-ray. X-ray is not deflected by electric and magnetic fields. Wavelength of a continuous X-ray depends on potential difference across the tuve. Wavelength of charasteristic X-ray depends on the atomic number.
Q. If frequency `(upsilongtupsilon_0)` of incident light becomes n times the initial frequency (v), then KE of the emitted photoelectrons becomes (`v_0` threshold frequency).

Options

A. n times of the initial kinetic energy
B. more than n times of the intial kinetic energy
C. less than n times of the initial kinetic energy
D. kinetic energy of the emitted photoelectrons remains unchanged

Explanation

To solve the problem regarding the photoelectric effect and the relationship between the frequency of incident light and the kinetic energy of emitted photoelectrons, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect Equation**:
The photoelectric effect can be described by the equation:
\[ E = h\nu = \phi + KE \]
where \(E\) is the energy of the incident photon, \(h\) is Planck's constant, \(\nu\) is the frequency of the incident light, \(\phi\) is the work function of the metal, and \(KE\) is the kinetic energy of the emitted photoelectron.

2. **Initial Frequency and Kinetic Energy**:
Let the initial frequency of the incident light be \(\nu\). The corresponding kinetic energy of the emitted photoelectrons can be expressed as:
\[ KE = h\nu - \phi \]

3. **New Frequency**:
If the frequency of the incident light is increased to \(n\) times the initial frequency, then the new frequency can be expressed as:
\[ \nu' = n\nu \]

4. **Kinetic Energy at New Frequency**:
The kinetic energy of the emitted photoelectrons at this new frequency will be:
\[ KE' = h\nu' - \phi = h(n\nu) - \phi = nh\nu - \phi \]

5. **Relating the Kinetic Energies**:
Now, we can relate the new kinetic energy \(KE'\) to the initial kinetic energy \(KE\):
\[ KE' = nh\nu - \phi \]
We can substitute \(\phi\) from the initial kinetic energy equation:
\[ \phi = h\nu - KE \]
Thus,
\[ KE' = nh\nu - (h\nu - KE) = nh\nu - h\nu + KE = (n-1)h\nu + KE \]

6. **Conclusion**:
This shows that the new kinetic energy \(KE'\) is greater than the initial kinetic energy \(KE\) by \((n-1)h\nu\). Hence, we can conclude that:
\[ KE' > n \cdot KE \]
Therefore, the kinetic energy of the emitted photoelectrons when the frequency is increased to \(n\) times the initial frequency is greater than \(n\) times the initial kinetic energy.

### Final Answer:
The kinetic energy of the emitted photoelectrons becomes greater than \(n\) times the initial kinetic energy.

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