Light of wavelength 4000A0 falls on a photosensitive metal and a negative 2 V potential stops the emitted electrons. The work function of the material (in eV) is approximately h=6.6×10-34Js, e=1.6×10-19C, c=3×108 ms-1

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Light of wavelength 4000 $\overset{0}{A}$ falls on a photosensitive metal and a negative 2 V potential stops the emitted electrons. The work function of the material (in eV) is approximately $(h = 6.6 \times 10^{- 34} Js, e = 1.6 \times 10^{- 19} C, c = 3 \times 10^{8} {ms}^{- 1})$

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

1.1

3.1

2.0

2.2

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} Energy of incident light E (eV) = \frac{12375}{4000} = 3 .09 eV \\ Stopping potential is - 2 V so K_{\max} = 2 eV \\ Hence by using E = W_{o} + K_{\max} \Rightarrow W_{o} = 1 .09 eV \approx 1 .1 eV \end{array}$