Email
Chat with tutor
logo

Ask Questions, Get Answers

X
 
Answer
Comment
Share
Q)

Light of wavelength $0.6\: \mu m$ from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is $0.5\: V$ and with light of wavelength $0.4\: \mu m$ from a sodium lamp, the stopping potential is $1.5\: V$. With this data, the value of $ \large\frac{h}{e}$ will be

$\begin {array} {1 1} (a)\;4 \times 10^{-19}\: Vs & \quad (b)\;0.25 \times 10^{-15}\: Vs \\ (c)\;4 \times 10^{-15}\: Vs & \quad (d)\;4 \times 10^{-8}\: Vs \end {array}$

 

1 Answer

Comment
A)
Ans : (c)
$hv =\phi + eV \: \: \: \: v = $frequency of radiation, $V$ = stopping potential
So, $hv_1 = \phi + eV = \phi + 0.5e$…………..(i)
$hv_2 = \phi + eV = \phi + 1.5e$……………..(ii)
(i)-(ii),
$h(v_1 – v_2) = e$
$ \Rightarrow hc\bigg( \large\frac{1}{\lambda_1} – \large\frac{1}{\lambda_2} \bigg) = e$
 
$ \Rightarrow \large\frac{h}{e} = \large\frac{\bigg[\large\frac{\lambda_1 \: \lambda_2}{( \lambda_2 – \lambda_1)} \bigg]}{c} = 4 \times 10^{-15} Vs$

 

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
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.
...