Question

The work function for metallic calcium is 2.9 eV. Calculate the kinetic energy in eV for electrons ejected from a metallic calcium surface by light of wavelength 405 nm.

Answer 1

239.73 km/s

Step-by-step explanation:

Given that:

The work function of the magnesium = 2.9 eV

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

1 eV = 1.6022 × 10⁻¹⁹ J

So, work function =
`2.9* 1.6022* 10^(-19)\ J=4.64638* 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=405\ nm=405* 10^(-9)\ m`

Thus, applying values as:

`(6.626* 10^(-34)* 3* 10^8)/(405* 10^(-9))=4.64638* 10^(-19)+(1)/(2)* 9.11* 10^(-31)* v^2`

`4.64638* \:10^(-19)+(1)/(2)* \:9.11* \:10^(-31)v^2=(6.626* \:10^(-34)* \:3* \:10^8)/(405* \:10^(-9))`

`(1)/(2)* 9.11* 10^(-31)* v^2=4.90814815* 10^(-19)-4.64638* 10^(-19)`

`(1)/(2)* 9.11* 10^(-31)* v^2=0.26176815* 10^(-19)`

`v^2=0.0574683* 10^(12)`

v = 2.3973 × 10⁵ m/s

Also, 1 m = 0.001 km

So, v = 239.73 km/s