The work function of Cs is 2.14 ev find the wavelength of the incident light if the stopping potential is 0.6 V.

Options

A. 326 nm
B. 454 nm
C. 524 nm
D. 232 nm

Explanation

To solve the problem step by step, we will use the concepts of the photoelectric effect and the relevant equations.

### Step 1: Write down the given data
- Work function (φ) of Cs = 2.14 eV
- Stopping potential (V₀) = 0.6 V

### Step 2: Understand the relationship between stopping potential, work function, and energy of incident light
According to Einstein's photoelectric equation:
\[ eV_0 = E - \phi \]
where:
- \( e \) is the charge of an electron (1 eV = 1.6 x 10⁻¹⁹ J)
- \( E \) is the energy of the incident photon
- \( \phi \) is the work function

### Step 3: Express the energy of the incident photon
The energy of the incident photon can also be expressed in terms of its wavelength (λ):
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (approximately \( 4.14 \times 10^{-15} \) eV·s)
- \( c \) is the speed of light (approximately \( 3 \times 10^8 \) m/s)

### Step 4: Substitute the expressions into the equation
From the photoelectric equation:
\[ eV_0 = \frac{hc}{\lambda} - \phi \]

Rearranging gives:
\[ \frac{hc}{\lambda} = eV_0 + \phi \]

### Step 5: Solve for λ
Now, we can solve for λ:
\[ \lambda = \frac{hc}{eV_0 + \phi} \]

### Step 6: Substitute the known values
We need to convert the constants into compatible units. Using:
- \( h = 4.14 \times 10^{-15} \) eV·s
- \( c = 3 \times 10^8 \) m/s
- \( eV_0 = 0.6 \) V
- \( \phi = 2.14 \) eV

Now substituting these values:
\[ \lambda = \frac{(4.14 \times 10^{-15} \text{ eV·s})(3 \times 10^8 \text{ m/s})}{0.6 + 2.14} \]

### Step 7: Calculate the denominator
\[ 0.6 + 2.14 = 2.74 \text{ eV} \]

### Step 8: Calculate λ
Now substituting back into the equation:
\[ \lambda = \frac{(4.14 \times 10^{-15})(3 \times 10^8)}{2.74} \]

Calculating the numerator:
\[ 4.14 \times 3 = 12.42 \]
So:
\[ \lambda = \frac{12.42 \times 10^{-7}}{2.74} \]

Calculating λ:
\[ \lambda \approx 454 \text{ nm} \]

### Final Answer
The wavelength of the incident light is approximately **454 nm**.

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