In a photoelectric experiment, the wavelength of the incident light is decreased from `6000A` to `4000A`. While the intensity of radiations remains the same,

Options

A. the cut off potential will decrease
B. the cut off potential will increase
C. the photoelectric current will increase
D. the kinetic energy of the emitted electrons will increase

Explanation

To solve the problem step by step, let's analyze the effects of decreasing the wavelength of incident light on the photoelectric effect.

### Step 1: Understand the relationship between wavelength and energy
The energy of a photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \( c \) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \( \lambda \) is the wavelength of the light.

### Step 2: Analyze the change in wavelength
Initially, the wavelength is \( \lambda_1 = 6000 \, \text{Å} \) (or \( 6000 \times 10^{-10} \, \text{m} \)), and it is decreased to \( \lambda_2 = 4000 \, \text{Å} \) (or \( 4000 \times 10^{-10} \, \text{m} \)).

As the wavelength decreases, the energy of the photons increases:
- For \( \lambda_1 = 6000 \, \text{Å} \):
\[ E_1 = \frac{hc}{6000 \times 10^{-10}} \]

- For \( \lambda_2 = 4000 \, \text{Å} \):
\[ E_2 = \frac{hc}{4000 \times 10^{-10}} \]

Since \( \lambda_2 < \lambda_1 \), it follows that \( E_2 > E_1 \).

### Step 3: Determine the effect on kinetic energy
The maximum kinetic energy (\( E_k \)) of the emitted electrons can be expressed as:
\[ E_k = E - W \]
where \( W \) is the work function of the material (a constant).

As the energy of the incident photons increases (from \( E_1 \) to \( E_2 \)), the maximum kinetic energy of the emitted electrons will also increase:
\[ E_{k2} = E_2 - W \]
\[ E_{k1} = E_1 - W \]
Since \( E_2 > E_1 \), it follows that \( E_{k2} > E_{k1} \).

### Step 4: Analyze the cutoff potential
The cutoff potential (or stopping potential, \( V_s \)) is related to the maximum kinetic energy of the emitted electrons:
\[ E_k = eV_s \]
where \( e \) is the charge of an electron.

Since the maximum kinetic energy increases, the cutoff potential must also increase:
\[ V_{s2} = \frac{E_{k2}}{e} \]
\[ V_{s1} = \frac{E_{k1}}{e} \]
Thus, \( V_{s2} > V_{s1} \).

### Step 5: Determine the effect on photoelectric current
The intensity of the radiation remains constant, which means the number of photons striking the surface per unit time remains the same. However, since the energy of each photon has increased, the number of emitted electrons may not change significantly. The photoelectric current depends on the number of emitted electrons, which can remain constant if the intensity is unchanged.

### Summary of Results
1. **Cutoff Potential**: Increases
2. **Kinetic Energy of Emitted Electrons**: Increases
3. **Photoelectric Current**: Remains the same (assuming intensity is constant)

### Final Answer
- The cutoff potential increases.
- The kinetic energy of the emitted electrons increases.
- The photoelectric current remains the same.