Dual Nature Of Light Question 355
Question: When a beam of 10.6 eV photons of intensity $ 2.0W/m^{2}, $ falls on a platinum surface of area $ 1.0\times {{10}^{-4}}m^{2} $ and work function 5.6 eV, 0.53% of the incident photons eject photoelectrons, then the number of photoelectrons emitted per second and their minimum & maximum energies (in eV) [Take $ 1eV=1.6\times {{10}^{-19}}J $ ] are respectively.
Options:
A) $ 1.18\times 10^{10}\ \text{eV}, 2\ \text{eV}, 5\ \text{eV} $
B) $ 1.18\times 10^{14},\text{eV},,0,\text{eV},,5,\text{eV}$
C) $ 2.18\times 10^{13}\ \text{eV}, 0\ \text{eV}, 5\ \text{eV}$
D) $ 3.11\times 10^{11},1eV,5eV $
Show Answer
Answer:
Correct Answer: B
Solution:
No. of photons/sec $ \text{=}\frac{\text{Energy incident on platinum surface per second}}{\text{Energy of one photon}} $ No. of photons incident per second $ =\frac{2\times 10\times {{10}^{-4}}}{10.6\times 1.6\times {{10}^{-19}}}=1.18\times 10^{14} $ As 0.53% of incident photons can eject photoelectrons $ \therefore $ No. of photoelectrons ejected per second $ =1.18\times 10^{14}\times \frac{0.53}{100}=6.25\times 10^{11} $ As 0.53% of incident photon can eject photoelectrons Minimum energy $ =0eV, $ Maximum energy $ =( 10.6-5.6 )eV=5eV $
BETA
