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

v = 301.62 km/s

Step-by-step 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* 1.6022* 10^(-19)\ J=5.8961* 10^(-19)\ J`

Using the equation for photoelectric effect as:

`E=\psi _0+\frac {1}{2}* m* v^2`

Also,
`E=\frac {h* c}{\lambda}`

Applying the equation as:

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

Where,

h is Plank's constant having value
`6.626* 10^(-34)\ Js`

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

m is the mass of electron having value
`9.11* 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* 10^(-9)\ m`

Thus, applying values as:

`(6.626* 10^(-34)* 3* 10^8)/(315* 10^(-9))=5.8961* 10^(-19)+(1)/(2)* 9.11* 10^(-31)* v^2`

`5.8961* \:10^(-19)+(1)/(2)* \:9.11* \:10^(-31)v^2=(6.626* \:10^(-34)* \:3* \:10^8)/(315* \:10^(-9))`

`(1)/(2)* 9.11* 10^(-31)* v^2=6.31047619* 10^(-19)-5.8961* 10^(-19)`

`(1)/(2)* 9.11* 10^(-31)* v^2=0.41437619* 10^(-19)`

`v^2=0.0909717212* 10^(12)`

v = 3.0162 × 10⁵ m/s

Also, 1 m = 0.001 km

So, v = 301.62 km/s