A metal surface is exposed to 500 nm radiation. The threshold frequency of the metal for photoelectric current is $$4.3 \times 10^{14}$$ Hz. The velocity of ejected electron is _________ $$\times 10^5$$ ms$$^{-1}$$ (Nearest integer)
[Use: h = $$6.63 \times 10^{-34}$$ Js, m$$_e$$ = $$9.0 \times 10^{-31}$$ kg]

We have been told that monochromatic radiation of wavelength $$\lambda = 500\ \text{nm}$$ (nanometres) is incident on a metal surface. The corresponding photon frequency is obtained from the fundamental relation between wavelength, frequency and the speed of light, namely $$c = \nu \lambda$$, where $$c = 3.0 \times 10^{8}\ \text{m s}^{-1}$$ is the speed of light.

Re-arranging, the frequency of the incident photons is

$$ \nu \;=\; \frac{c}{\lambda}. $$

Substituting $$c = 3.0 \times 10^{8}\ \text{m s}^{-1}$$ and $$\lambda = 500\ \text{nm} = 500 \times 10^{-9}\ \text{m} = 5.0 \times 10^{-7}\ \text{m},$$ we get

$$ \nu \;=\; \frac{3.0 \times 10^{8}}{5.0 \times 10^{-7}} \;=\; 6.0 \times 10^{14}\ \text{Hz}. $$

The metal has a threshold (cut-off) frequency $$\nu_0 = 4.3 \times 10^{14}\ \text{Hz}$$. According to Einstein’s photoelectric equation, the maximum kinetic energy $$K_{\text{max}}$$ of an ejected electron is given by

$$ K_{\text{max}} = h(\nu - \nu_0), $$

where $$h = 6.63 \times 10^{-34}\ \text{J s}$$ is Planck’s constant.

First find the excess frequency above the threshold:

$$ \nu - \nu_0 = 6.0 \times 10^{14} - 4.3 \times 10^{14} = 1.7 \times 10^{14}\ \text{Hz}. $$

Now calculate the kinetic energy:

$$ K_{\text{max}} = h(\nu - \nu_0) = (6.63 \times 10^{-34}\ \text{J s})(1.7 \times 10^{14}\ \text{Hz}) = 1.1271 \times 10^{-19}\ \text{J}. $$

For an electron, this kinetic energy is related to its speed $$v$$ through the classical formula $$K = \tfrac12 m_e v^{2}$$, where the electron mass is $$m_e = 9.0 \times 10^{-31}\ \text{kg}$$. Stating the formula first:

$$ \frac12 m_e v^{2} = K_{\text{max}}. $$

Solving for $$v$$ gives

$$ v = \sqrt{\frac{2K_{\text{max}}}{m_e}}. $$

Substituting the numerical values:

$$ v = \sqrt{\frac{2(1.1271 \times 10^{-19})}{9.0 \times 10^{-31}}} = \sqrt{\frac{2.2542 \times 10^{-19}}{9.0 \times 10^{-31}}}. $$

Divide the mantissas and handle the powers of ten separately:

$$ \frac{2.2542}{9.0} = 0.250466\quad\text{and}\quad 10^{-19} \div 10^{-31} = 10^{12}. $$

Thus

$$ v = \sqrt{0.250466 \times 10^{12}} = \sqrt{2.50466 \times 10^{11}}. $$

Taking the square root:

$$ v \approx 5.0 \times 10^{5}\ \text{m s}^{-1}. $$

Because the question asks for the velocity expressed as “_________ $$\times 10^{5}$$ m s$$^{-1}$$ (nearest integer)”, we read off the factor as $$5$$.

So, the answer is $$5$$.