Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

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
`        An X-ray tube operates at 40 kV. Suppose the electron converts 70% of its energy into a photon at each collision. Find the lowest three wavelength emitted from the tube. Neglect the energy imparted to the atom with which the electron collides. `
3 years ago

Jitender Pal
365 Points
```										V = 40 KV = 40 * 10^3 V
Energy = 40 * 10^3 eV
Energy utilized = 70/100 *40 * 10^3 = 28 * 10^3 eV
11. V = 40 KV = 40 * 10^3 V
Energy = 40 * 10^3 eV
Energy utilized = 70/100 * 40 * 10^3 = 28 * 10^3 eV
λ = hc/E = 1242 – ev nm/ 28* 10^3 ev ⇒ 444.35 * 10^-3 nm = 44.35 pm.
For other wavelengths,
E = 70% (left over energy) = 70/100 * (40 - 28)10^3 = 84 * 10^2.
λ’ = hc/E = 1242/ 25.2* 10^2 = 49.2857 * 10^-2 nm = 493 pm.

```
3 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Modern Physics

View all Questions »
• 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