Question

The work function for metallic magnesium is 3.68 eV. Calculate the velocity in km/s for electrons ejected from a metallic magnesium surface by light of wavelength 315 nm.

Answer 1

Answer:

v = 301.62 km/s

Explanation:

Given that:

The work function of the magnesium = 3.68 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, work function = $3.68\times 1.6022\times 10^{-19}\ J=5.8961\times 10^{-19}\ J$

Using the equation for photoelectric effect as:

$E=\psi _0+\frac {1}{2}\times m\times v^2$

Also, $E=\frac {h\times c}{\lambda}$

Applying the equation as:

$\frac {h\times c}{\lambda}=\psi _0+\frac {1}{2}\times m\times v^2$

Where,  

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

m is the mass of electron having value $9.11\times 10^{-31}\ kg$

$\lambda$ is the wavelength of the light being bombarded

$\psi _0=Work\ function$

v is the velocity of electron

Given, $\lambda=315\ nm=315\times 10^{-9}\ m$

Thus, applying values as:

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{315\times 10^{-9}}=5.8961\times 10^{-19}+\frac{1}{2}\times 9.11\times 10^{-31}\times v^2$

$5.8961\times \:10^{-19}+\frac{1}{2}\times \:9.11\times \:10^{-31}v^2=\frac{6.626\times \:10^{-34}\times \:3\times \:10^8}{315\times \:10^{-9}}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=6.31047619\times 10^{-19}-5.8961\times 10^{-19}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=0.41437619\times 10^{-19}$

$v^2=0.0909717212\times 10^{12}$

v = 3.0162 × 10⁵ m/s

Also, 1 m = 0.001 km

So, v = 301.62 km/s