Light of wavelength A which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with same velocity, then stopping potential will

Options

A. increase
B. decrease
C. be zero
D. become exactly half

Answer

To solve the problem step by step, we will analyze the relationship between the wavelength of light, the energy of the incident photons, and the stopping potential.

### Step 1: Understand the relationship between wavelength and photon energy
The energy of a photon is given by the equation:
\[ E = h \nu \]
where \( E \) is the energy, \( h \) is Planck's constant, and \( \nu \) is the frequency of the light. The frequency \( \nu \) can be expressed in terms of wavelength \( \lambda \) as:
\[ \nu = \frac{c}{\lambda} \]
where \( c \) is the speed of light. Therefore, we can rewrite the energy equation as:
\[ E = \frac{hc}{\lambda} \]

### Step 2: Analyze the effect of changing the wavelength
Since the energy \( E \) is inversely proportional to the wavelength \( \lambda \), if we decrease the wavelength (i.e., \( \lambda \) becomes smaller), the energy \( E \) will increase.

### Step 3: Relate photon energy to kinetic energy and stopping potential
The energy of the incident photon is used to overcome the work function \( \phi \) of the material and provide kinetic energy to the emitted photoelectrons. This relationship can be expressed as:
\[ E = K.E + \phi \]
where \( K.E \) is the maximum kinetic energy of the emitted electrons. The stopping potential \( V_0 \) is related to the kinetic energy of the electrons by:
\[ K.E = e V_0 \]
where \( e \) is the charge of the electron. Therefore, we can write:
\[ \frac{hc}{\lambda} = e V_0 + \phi \]

### Step 4: Determine the effect on stopping potential when wavelength decreases
If the wavelength is decreased, the energy \( E \) increases. Since the work function \( \phi \) remains constant for a given material, an increase in energy \( E \) will result in an increase in the maximum kinetic energy \( K.E \) of the emitted electrons. Consequently, since \( K.E = e V_0 \), the stopping potential \( V_0 \) must also increase.

### Conclusion
Thus, when the incident wavelength is decreased while keeping the emitted photoelectrons moving with the same velocity, the stopping potential will increase.

### Final Answer
The stopping potential will **increase**.