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The photoelectric work - function of po...

The photoelectric work - function of potassium is 2.3 eV. If light
having a wavelength of `2800 Å` falls on potassium, find
(a) the kinetic energy in electron volts of the most energetic electrons ejected.
(b) the stopping potential in volts.

A

2.1 eV , 2.1V

B

2.3 eV , 2.4V

C

4.2 eV , 4.1V

D

4.5 eV , 2.7V

Text Solution

Verified by Experts

The correct Answer is:
A

Given, `W=2.3eV,lamda=2800`Å
`therefore` E(in eV)=`(12375)/(lamda("in Å"))`
`=(12375)/(2800)`
`=4.4eV`
`K_(max)=E-W`
`=(4.4-2.3)eV=2.1eV`
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