A silver sphere of radius 1cm and work function $4.7 e V$ is suspended from an insulating thread in free-space. It is under continuous illumination of $200 n m$ wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is $A \times 10^{Z}$ (Where $1 < A < 10$ ). The value of $Z$ is _________

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

answer is 7.

(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

Photo emission will stop when potential on silver sphere becomes equal to the stopping potential.
$∴ \frac{h c}{λ} - W = e V_{0}$
Here, $V_{0} = \frac{1}{4 π \in_{0}} \frac{z e}{r}$ $∴ \frac{h c}{λ} - W = \frac{1}{4 π \in_{0}} \frac{n e^{2}}{r}$ $∴ (\frac{1240}{200} e V) - (4.7 e V)$

$= \frac{9 \times 10^{9} \times n \times 1.6 \times 10^{- 19}}{10^{- 2}}$ $(6.2 - 4.7) \times 1.6 \times 10^{- 19}$ $= \frac{9 \times 10^{9} \times n \times {(1.6 \times 10^{- 19})}^{2}}{10^{- 2}}$ $o r n = \frac{1.5 \times 10^{- 2}}{9 \times 1.6 \times 10^{- 10}} = 1.04 \times 10^{7} = 7$

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET