Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited state to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the materials is:

a) 

4\times 10^{15}\: \: Hz

b) 

5\times 10^{15}\: \: Hz

c) 

1.6\times 10^{15}\: \: Hz

d) 

2.5 \times 10^{15}\: \: Hz

please explain

 

Answers (1)

Energy released from emission of electron

E=10.2eV

From photo electric equation work function,

\begin{array}{l} \phi=\mathrm{E}-\mathrm{eV}=\mathrm{hv} \\ \mathrm{v}=\frac{\mathrm{E}-\mathrm{eV}}{\mathrm{h}} \\ \\ =\frac{(10.2-3.57) \mathrm{e}}{6.67 \times 10^{-34}} \\ \\ =\frac{6.63 \times 1.6 \times 10^{-19}}{6.67 \times 10^{-34}}=1.6 \times 10^{15} \mathrm{Hz} \end{array}

Posted by

Satyajeet Kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’
anam-khan

NIT Puducherry Cut-off 2024: Computer Science and Engineering had highest closing rank last year
anam-khan

What after JEE Main results 2024? All you need to know about JEE Adv, CSAB, JoSAA counselling

Latest Question