Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The energy that should be added to an electron to reduce its de Broglie wavelength from 1 nm to 0.5 nm is:  Option: 1  four times th

The energy that should be added to an electron to reduce its de Broglie wavelength from 1 nm to 0.5 nm is:
 

Option: 1

 four times the initial energy
 


Option: 2

 equal to the initial energy


Option: 3

 twice the initial energy

 


Option: 4

 thrice the initial energy


Answers (1)

best_answer

de Broglie wavelength,
\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}\quad \quad \quad (i)

After decreasing wavelength,
\lambda^{\prime}=\frac{\mathrm{h}}{\mathrm{p}^{\prime}}=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mE}^{\prime}}} \quad \quad \quad (ii)

From eqs. (i) and (ii),
\frac{\lambda^{\prime}}{\lambda}=\sqrt{\frac{E}{E^{\prime}}}

Putting values of \lambda^{\prime} and \lambda, we get
\begin{aligned} & \frac{\mathrm{E}}{\mathrm{E}^{\prime}}=\left(\frac{0.5}{1}\right)^2 \\ & \therefore \mathrm{E}^{\prime}=\frac{\mathrm{E}}{0.25}=4 \mathrm{E} \end{aligned}

The energy should be added to decrease the wavelength=E^{\prime}-E=4 E-E=3 E

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

Trending Articles/News

anam-khan

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions
anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in
anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question