Questions
In a photoelectric effect experiment a metallic surface having work function 2.2eV is illuminated with a light of wavelength . An incident photon makes a collision with a free electron that lies well inside the target metal and the electron makes one collision before being ejected from the surface. If 10% of the energy of the electron is lost in the collision, The kinetic energy of the ejected electron is
detailed solution
Correct option is D
Energy of photon =hcλ=124204000ev⇒hv=3.1evEnergy lost in the collision =10100 x 3.105 ev=0.31 evNow, hv=ϕ+k+Lost energy⇒ 3.1 = 2.2 + KE + 0.31 ⇒KE=0.59 evTalk to our academic expert!
Similar Questions
Light described at a place by equation falls on metal surface having work function 2 eV, the maximum kinetic energy of photoelectrons?
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests