Email
Chat with tutor
logo

Ask Questions, Get Answers

X
 
Answer
Comment
Share
Q)

When light of wavelength 300 nm falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, light of wavelength 600 nm is sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is

$\begin {array} {1 1} (a)\;1:2 & \quad (b)\;2:1 \\ (c)\;4:1 & \quad (d)\;1:4 \end {array}$

 

1 Answer

Comment
A)
Ans : (b)
$ \phi_1 = \large\frac{hc}{\lambda h_1} = \large\frac{hc}{300}$
$ \phi_2 = \large\frac{hc}{\lambda h_2} = \large\frac{hc}{600}$
So, $ \large\frac{ \phi_1}{ \phi_2} = \large\frac{600}{300} = \large\frac{2}{1}$

 

Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Continue
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Continue
...