MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Simran Bhatia Grade: 11
        Two neutral particles are kept 1 m apart. Suppose by some mechanism some charge is transferred from one particle to the other and the electric potential energy lost is completely converted into a photon. Calculate the longest and the next smaller wavelength of the photon possible.
3 years ago

Answers : (2)

Navjyot Kalra
askIITians Faculty
654 Points
										Sol. r = 1 m

Energy = kq^2/R = kq^2/1
Now, kQ^2/1 = hc/λ or λ = hc/kq^2
For max ‘λ’, ‘q’ should be min,
For λ hc/kq^2 – 0.863 * 10^3 = 863 m.
For next smaller wavelength = 6.63 * 3 *10^-34 * 10^8/9 * 10^9 *(1.6 * 2)^2 * 10^-38 = 863/4 = 215.74 m
3 years ago
Dhawal Patil
24 Points
										
As charge is quantized, the least amount of charge that can be transferred is e.
According to the question, energy lost is completely transferred in the form of the photon.
Please note that the energy is lost as conservative forces are considered to be negative if they are attractive in nature.
As electric potential energy  =
 -\frac{1}{4\pi\epsilon_o}\!\!\frac{{q}^2}{r}
And energy of the photon = h\nu
 
We have :
-\frac{1}{4\pi\epsilon_o}\!\!\frac{{q}^2}{r} = h\nu
Sustituting, \nu = \frac{c}{\lambda} and discarding the minus sign, we get:
\frac{1}{4\pi\epsilon_o}\!\!\frac{q^2}{r} = \frac{hc}{\lambda}
which implies,
 \lambda=\frac{hc}{\frac{1}{4\pi\epsilon_o}\!\!\frac{q^2}{r}}
Also, \lambda_{n^{th}largest} = \frac{(6\!\cdot\!63.10^{-34}Js).(3\!\cdot\!0.10^8ms^{-1}).1m)}{(9.10^9Nm^2C^{-2}).(n.1\!\cdot\!6.10^{-19})^2}
\therefore\lambda_{largest} = \frac{(6\!\cdot\!63.10^{-34}Js).(3\!\cdot\!0.10^8ms^{-1}).1m)}{(9.10^9Nm^2C^{-2}).(1\!\cdot\!6.10^{-19})^2}\therefore\lambda_{largest} = \frac{(6\!\cdot\!63.10^{-34}Js).(3\!\cdot\!0.10^8ms^{-1}).1m)}{(9.10^9Nm^2C^{-2}).(2.1\!\cdot\!6.10^{-19})^2}=\frac{\lambda_{largest}}{4}
10 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: Rs. 2,756
  • View Details
Get extra Rs. 413 off
USE CODE: COUPON10
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details
Get extra Rs. 445 off
USE CODE: COUPON10

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details