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 potential difference across the tube is increased then

Options

A. `lamda_(min)` will decrease
B. characteristic wavelength will increase
C. `lamda_(min)` will increase
D. none of these

Explanation

To solve the question regarding the effect of increasing the potential difference across an X-ray tube, we can follow these steps:

### Step 1: Understand the relationship between potential difference and X-ray production
When electrons are accelerated through a potential difference (V) in an X-ray tube, they gain kinetic energy equal to the work done on them by the electric field. This kinetic energy is given by the equation:
\[ KE = eV \]
where \( e \) is the charge of the electron.

### Step 2: Relate kinetic energy to X-ray wavelength
The energy of the emitted X-rays can be expressed in terms of their wavelength (\( \lambda \)) using the equation:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

### Step 3: Set up the relationship between potential difference and wavelength
From the above equations, we can equate the kinetic energy of the electrons to the energy of the emitted X-rays:
\[ eV = \frac{hc}{\lambda} \]
Rearranging this gives us:
\[ \lambda = \frac{hc}{eV} \]

### Step 4: Analyze the effect of increasing potential difference
From the equation \( \lambda = \frac{hc}{eV} \), we can see that as the potential difference \( V \) increases, the wavelength \( \lambda \) decreases. This means that the minimum wavelength of the emitted X-rays becomes shorter.

### Conclusion
Therefore, if the potential difference across the tube is increased, the minimum wavelength of the emitted X-rays decreases.

### Final Answer
The correct conclusion is that if the potential difference across the tube is increased, the minimum wavelength of the emitted X-rays will decrease.

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