#### Thank you for registering.

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

Click to Chat

1800-5470-145

+91 7353221155

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

# CHEMISTRY For 3s orbital of hydrogen atom, the normalised wave function is given by y3s = {1/[81(3 pie)^1/2} * (1/a)^3/2 * [27 - (18r)/a + (r)^2/(a)^2] * e^r/3a The above mentioned orbital (3s) has two nodes at 1.9a0 and xa0. What is the approximate value of x? (a) 9 (b) 7 (c) 6 (d) 2

SAGAR SINGH - IIT DELHI
879 Points
10 years ago

Dear student,

To get the value of x, put the normalised wave function to be zero. This is just a hint, u can solve the equation on ur own..

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Sagar Singh

B.Tech, IIT Delhi

sagarsingh24.iitd@gmail.com

Pankaj Rajpal
13 Points
3 years ago
A node is where it goes to zero; BTW it must be e^(MINUS r/3ao) otherwise it blows up

in this case ignore everything but [ 27- 18r/a0 + 2r^2/a0^2]
By solving this quadratic equation we get the two answer from which we can find the values of nodes

1.90

7.10