Q.1.A beam of electrons accelerated with 4.64 V was passed through a tube having mercury vapours .As a result of absorption,electronic changes occured with mercury atoms and light was emitted.If the full energy of the single electron ws converted into light ,what was the wave number of emitted light.
Q.2.The binding energy of an electron in the ground state of an atom is equal to 24.6eV.Find the enegry required to remove both electrons from the atom.

Question

askiitian.expert- chandra sekhar · Answer

Hi Vinay,

1) 4.64 eV is the energy of the electron

if its total energy is converted photons then

hc/λ=4.64*1.602*10^-19J

1/λ=3.74*10⁶

2)B.E is 24.6 eV

for hydrogen it is 13.6 eV

24.6 which is nearly twice that of hydrogen implies the atom is He

for first electron it is 24.6 eV

for second electron it is 27.2 eV

for removing both the electrons it is 27.2+24.6=51.8 eV

Sankalp Samuel Das · Answer

A.1) V= 4.64V E= QV = (1.6 x 10 -19 x 4.64) J = 7.424 x 10 -19 J E= hc /&lambda; = hcv , [where v= 1/&lambda;, v=wave number] => v= E/hc => v= (7.424 x 10 -19 ) / (6.62 x 10 -34 x 3 x 10 8) m -1 => v= 37400 c m -1

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

2 Answers

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Physical Chemistry

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship