Question

The threshold frequency for photoelectric emission of electrons from platinum is $1.3\times {10}^{15}{s}^{-1}$. Then incorrect statement is:

• A. The number of electrons ejected from the metal surface will depend upon the intensity
• B. The radiation having energy $1\times {10}^{-20}J$ will be able to cause photoelectric effect
• C. If radiation having frequency $1.3\times {10}^{14}{s}^{-1}$ strikes the platinum surface, photo electrons will not be ejected
• D. The kinetic energy of the emitted electrons will increase with increase in frequency provided it is greater than $1.3\times {10}^{15}{s}^{-1}$

Answer 1

Answer: (B)

The radiation having energy $1\times {10}^{-20}J$ will be able to cause photoelectric effect

Energy required for photoelectric emission
$=h\lambda$
$\lambda =13\times {10}^{15}/sec$
$\therefore 6.6\times {10}^{-34}\times 1.3\times {10}^{15}$
$=8.58\times {10}^{-19}J$
$\therefore$ B is correct
$1\times {20}^{-20}J<thresholdenergy$