Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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.
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
A.1)
V= 4.64V
E= QV
= (1.6 x 10-19 x 4.64) J
= 7.424 x 10-19 J
E= hc /λ
= hcv , [where v= 1/λ, v=wave number]
=> v= E/hc
=> v= (7.424 x 10-19) / (6.62 x 10-34 x 3 x 108) m-1
=> v= 37400 cm-1
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !