Chat with tutor

Ask Questions, Get Answers


The vapours of Hg absorb some electrons accelerate by a potential difference of 4.5volts as a result of which light is emitted .If the full energy of single incident electron is supposed to be converted into light emitted by single Hg atom,find the wave number $\big(\large\frac{1}{\lambda}\big)$ of the light.

$\begin{array}{1 1}(a)\;2.63\times 10^6m^{-1}&(b)\;4.62\times 10^6m^{-1}\\(c)\;3.63\times 10^6m^{-1}&(d)\;3.63\times 10^2m^{-1}\end{array}$

1 Answer

Energy of an accelerated electron =$Q.V=1.602\times 10^{-19}\times 4.5$
$\Rightarrow 7.209\times 10^{-19}J$
This energy is completely converted into light.
$\large\frac{hc}{\lambda}$$=7.209\times 10^{-19}$
$\large\frac{6.625\times 10^{-34}\times 3.0\times 10^8}{\lambda}$$=7.209\times 10^{-19}$
$\large\frac{1}{\lambda}$=wave no =$3.63\times 10^6m^{-1}$
Hence (c) is the correct answer.
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.
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
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.