Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • In an experiment on the photo-electric effect, singly ionized helium is excited electronically to different energy levels. The light emitted by ionised helium is incident on a photo-elect

In an experiment on the photo-electric effect, singly ionized helium is excited electronically to different energy levels. The light emitted by ionised helium is incident on a photo-electric plate in a photocell. When helium is excited to fourth energy level, then the observed “stopping potential” of the photocell is found to be five times the stopping potential measured when the photoelectrons are produced by using light emitted by hydrogen atom, excited to the third energy level. The work function of then material of the photo-electric plate is:

Option: 1

3.96 \mathrm{eV}


Option: 2

2.36 \mathrm{eV}


Option: 3

6.8 \mathrm{eV}


Option: 4

2.55 \mathrm{eV}


Answers (1)

best_answer

Energy of the photon with the highest frequency in the emission spectrum is
For helium (excited to the 4^{\text {th }} level) : \mathrm{f}_1=(2)^2 \mathrm{R} \mathrm{c}\left[\frac{1}{1^2}-\frac{1}{4^2}\right]
For hydrogen (excited to the 3^{\text {rd }} level) : \mathrm{f}_2=\mathrm{Rc}\left[\frac{1}{1^2}-\frac{1}{3^2}\right]
Using Einstein's equation of the photo-electric effect, we obtain,
\mathrm{hf}_1-\mathrm{W}=\mathrm{eV}_1 \\ \mathrm{hf}_2-\mathrm{W}=\mathrm{eV}_2 \\
where \mathrm{V}_1=5 \mathrm{~V}_2 (given)
Using (3), (4) and (5) we obtain,
\frac{\mathrm{hf}_1-\mathrm{W}}{\mathrm{e}}=5\left(\frac{\mathrm{hf}_2-\mathrm{W}}{\mathrm{e}}\right)
\Rightarrow 4 \mathrm{~W}=\mathrm{h}\left(5 \mathrm{f}_2-\mathrm{f}_1\right)
\Rightarrow \mathrm{W}=\frac{\mathrm{h}}{4}\left[5 \mathrm{f}_2-\mathrm{f}_1\right]
Putting the values of f_1  and f_2  from (1) and (2) in (6) we obtain,

\mathrm{W}=\frac{\mathrm{h}}{4}\left[\left(1-\frac{1}{9}\right) 5 \mathrm{Rc}-4\left(1-\frac{1}{16}\right) \mathrm{Rc}\right]=\frac{\mathrm{h}}{4} \mathrm{Rc}\left[\frac{8 \times 5}{9}-\frac{15}{4}\right]=0.694 \mathrm{Rhc}
Putting Rhc =13.6 \mathrm{eV} we obtain

\mathrm{W}=2.36 \mathrm{eV}

Posted by

Divya Prakash Singh

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Latest Question