Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
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

UV light of the wavelength 800 A and 700 A falls on the hydrogen atoms in their ground state and liberates electrons with KE 1.8 eV and 4 eV respectively. Calculate Planck's constant.

UV light of the wavelength 800 A and 700 A falls on the hydrogen atoms in their ground state and liberates electrons with KE 1.8 eV and 4 eV respectively. Calculate Planck's constant.

Grade:11

1 Answers

Ramesh V
70 Points
12 years ago

KE = hc/λ - Wf       where  Wf is work function

1.8 = hc / 8*10-8  - Wf       .       ... (1)

4 = hc / 7*10-8  - Wf                  ....(2)

2.2 = hc*108 [1/7 - 1/8]

Plancks constat : h = 4.106 x 10-15 eV-sec  or  6.578 x 10-34 J-sec

---

Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and
we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.

Regards,
Naga Ramesh
IIT Kgp - 2005 batch

 


 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free