Work function of nickel is 5.01 eV. When ultraviolet radiation of wavelength `200A` is incident of it, electrons are emitted. What will be the maximum velocity of emitted electrons?

Options

A. `3xx10^(8) ms^(-1)`
B. `6.46xx10^(5)ms^(-1)`
C. `10.36xx10^(5) ms^(-1)`
D. `8.54xx10^(6) ms^(-1)`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the energy of the incident ultraviolet radiation
The energy of the incident radiation can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(h\) (Planck's constant) = \(4.14 \times 10^{-15} \, \text{eV s}\)
- \(c\) (speed of light) = \(3 \times 10^8 \, \text{m/s}\)
- \(\lambda\) (wavelength) = \(200 \, \text{Å} = 200 \times 10^{-10} \, \text{m}\)

Substituting the values:
\[ E = \frac{(4.14 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{200 \times 10^{-10} \, \text{m}} = \frac{1.242 \times 10^{-6} \, \text{eV m}}{200 \times 10^{-10} \, \text{m}} = 6.21 \, \text{eV} \]

### Step 2: Calculate the kinetic energy of the emitted electrons
The maximum kinetic energy (\(KE_{max}\)) of the emitted electrons can be calculated using the equation:
\[ KE_{max} = E - \phi \]
where \(\phi\) is the work function. Given that the work function of nickel is \(5.01 \, \text{eV}\):
\[ KE_{max} = 6.21 \, \text{eV} - 5.01 \, \text{eV} = 1.20 \, \text{eV} \]

### Step 3: Convert kinetic energy to joules
To find the maximum velocity, we need to convert the kinetic energy from eV to joules. The conversion factor is:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Thus,
\[ KE_{max} = 1.20 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.92 \times 10^{-19} \, \text{J} \]

### Step 4: Use the kinetic energy formula to find maximum velocity
The kinetic energy is also given by the equation:
\[ KE = \frac{1}{2} mv^2 \]
where \(m\) is the mass of the electron (\(9.1 \times 10^{-31} \, \text{kg}\)). Rearranging for \(v\):
\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]
Substituting the values:
\[ v = \sqrt{\frac{2 \cdot 1.92 \times 10^{-19} \, \text{J}}{9.1 \times 10^{-31} \, \text{kg}}} \]
Calculating this gives:
\[ v = \sqrt{\frac{3.84 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx \sqrt{4.22 \times 10^{11}} \approx 6.5 \times 10^5 \, \text{m/s} \]

### Final Answer
The maximum velocity of the emitted electrons is approximately:
\[ v_{max} \approx 6.5 \times 10^5 \, \text{m/s} \]