use W=hVo and E = hc/(/lambda)
52) Threshold frequency of a photon required to eject a photoelectron from the surface of cesium is
Answer is:
4.3??1014Hz
Explanation:
Related Dual Nature of Radiation MCQ with Answers
Answer is:
12.4 eV
Explanation:
use W=hVo and E = hc/(/lambda)
Answer is:
2??105ms?1
Explanation:
Use 1/2mv2max=hv???
Answer is:
1.6??108cms?1
Explanation:
Use 1/2mv2max=hv???
Answer is:
2.4??10?12?erg
Explanation:
use W=hVo and E = hc/(/lambda)