Search Colleges, Courses, Exams, Questions and Articles

Ask a query on 8879100846 Ask Login | Sign Up

 

 

› Ask & Answer › Question

An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electron and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000 Å. What is the maximum kinetic energy of the emitted photoelectron?

Option 1 -

1.41 eV

Option 2 -

7.61 eV

Option 3 -

3.3 eV

Option 4 -

No photoelectron would be emitted

0 2 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    a month ago
    Correct Option - 1


    Detailed Solution:

    Energy of electron = 3 eV
    It forms H atom in n=3 state. Energy released E = 3 - (-13.6/9) = 3 + 1.51 = 4.51 eV.
    Photon Energy = 4.51 eV
    Threshold energy = hc/λ = 12400eVÅ / 4000Å = 3.1 eV.
    kE_max = 4.51 - 3.1 = 1.41 eV

     

Similar Questions for you

If work done in increasing the size of rectangular soap film with dimensions 8 cm × 3.75 cm to 10 cm × 6 cm is 2 × 10–4 J. The surface tension of film in newton per meter is

 
A
alok kumar singh

Change in surface energy = work done

 

A stationary hydrogen atom de excites from first excited state to ground state. Find recoil speed of hydrogen atom up to nearest integral value, (mass of hydrogen atom
= 1.8 × 10–27 kg)
A
alok kumar singh

| Δ E 0 | = ( − 1 3 6 { 1 − 1 4 } )   e V

|DE0| = –10.2

λ = 1 2 4 0 0 1 0 . 2 × 1 0 − 1 0   m

ρ = h λ = 6 . 6 3 × 1 0 − 3 4 × 1 0 . 2 1 2 4 0 0 × 1 0 − 1 0                  

? m v = h λ            

  1 . 8 × 1 0 − 2 7           

v = 6 . 6 3 × 1 0 . 2 × 1 0 − 3 4 1 2 4 0 0 × 1 0 − 1 0

v = 6 . 6 3 × 1 0 . 2 1 2 4 0 0 × 1 . 8 × 1 0 3            

= 6 . 6 3 × 1 0 2 1 2 4 × 1 . 8 = 3 . 0 2 ]

= 3 m/s

Electron in an hydrogen atom is excited to an energy level having energy –0.85 eV. Find the number of possible transitions it can make while deexcitation.
A
alok kumar singh

− 0 . 8 5 = − 1 3 . 6 n 2  

n = 4

Number of transitions = 4 × 3 2 = 6  

The ratio of kinetic energy & potential energy in 5th excited state of Hydrogen atom is
A
alok kumar singh

Kinetic energy: Potential energy = 1 : –2

If wavelength corresponding to 2nd line of Lyman series is l, then the wavelength corresponding to last line of Balmer series will be:
A
alok kumar singh

Kindly go through the solution

 

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 687k Reviews
  • 1800k Answers

Learn more about...

Share Your College Life Experience

Write a Review

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Ask Now
Community GuidelinesUser Points System
QuestionsDiscussionsTags
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.

Need guidance on career and education? Ask our experts

Characters 0/140

The Answer must contain atleast 20 characters.

Add more details

Characters 0/300

The Answer must contain atleast 20 characters.

Keep it short & simple. Type complete word. Avoid abusive language. Next

Your Question

Edit

Add relevant tags to get quick responses. Cancel Post