Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
0 votes

Light quanta with an energy 4.9 eV ejects photoelectrons from a metal with work function 4.5 eV .Then the maximum impulse transimitted to surface of metal when each electron flies out. Then de-Broglie wavelength of photoelctrons :

$(a)\;2 \times10^{-6} m\qquad(b)\;2 \times10^{-8} m\qquad(c)\;2 \times10^{-9} m\qquad(d)\;2 \times10^{-10} m$

Can you answer this question?

1 Answer

0 votes
Answer : (b) $\;2 \times10^{-8} m$
Explanation :
Einstein photoelectric equation
$\large\frac{1}{2} mV^2=4.9-4.5=0.4 eV=0.4\times16\times10^{-19 } J$
$V=\sqrt{\large\frac{2 \varepsilon}{M}}$
The momentum charge that is equal to impulse
$l=mV=m\;\sqrt{\large\frac{2 \varepsilon}{m}}\sqrt{2m \varepsilon}$
$I=\sqrt{2 \times 9.1\times10^{-31}\times0.4\times1.6\times10^{-19}}$
$I=mV=3.45 \times10^{-25}\;Kgm/second$
The de-Broglie wavelength $\;\lambda=\large\frac{h}{mV}$
$=\large\frac{6.63 \times 10^{-33}}{3.45 \times10^{-25}}=2 \times 10^{-8} m\;.$
answered Feb 26, 2014 by yamini.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App