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
```        The electric field associated with a light wave is given by
E = E base 0 sin [(1.57 × 10^7 m^-1) (x – ct)].
Find the stopping potential when this light is used in an experiment on photoelectric effect with the emitter having work function 1.9 eV.
```
3 years ago

Kevin Nash
332 Points
```										Sol. E = E base 0 sin[(1.57 * 10^7 m^-1) (x - ct)]
W = 1.57 * 10^7 * C
⇒ f = 1.57 * 10^7 *3 * 10^8/2π Hz W base 0 = 1.9 ev
Now eV base 0 = hv – W base 0
= 4.14 * 10^15 * 1.57 * 3 * 10^15/2π – 1.9 ev
= 3.105 – 1.9 = 1.205 ev
So, V base 0 = 1.205 * 1.6 *10^-19/1.6 * 10^-19 = 1.205 V.

```
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