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
Amit Saxena Grade: upto college level
`        A horizontal cesium plate (φ = 1.9 eV) is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of v so that the vertically upward component of velocity is non-positive for each photoelectron?`
3 years ago

396 Points
```										Sol. When λ  = 250 nm
Energy of photon = hc/λ = 1240/250 = 4.96 ev
∴ K.E. – hc/λ = 1240/250 = 4.96 ev
Velocity to be non positive for each photo electron
The minimum value of velocity of plate should be = velocity of photo electron
∴ Velocity of photo electron = √2Ke/m
= √3.06/9.1 * 10^-31 = √3.06 * 1.6 * 10^-19/9.1 * 10^-31 = 1.04 * 10^6 m/sec.

```
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
• Complete Physics Course - Class 11
• OFFERED PRICE: Rs. 2,968
• View Details