Question

I UITYuuy TUO
v v
re
3. Radius difference between 4th and 6th Bohr orbit of
H-atom is​

Answer 1

Answer:

$0.529 \times {n}^{2} \div z \\ 0.529( {6}^{2} - {4}^{2} ) \div 1 \\ 0.529(36 - 16) \\ \\ 0.529 \times 20 = ans$

Answer 2

Answer:

THE ANSWER TO THIS QUESTION IS 10.58 ARMSTRONG

Explanation:

• Using the "BOHR MODEL " we can find out the radius of any orbit

• Bohr said that an electron could revolve only in those orbits where its angular momentum is quantised i.e where the angular momentum is an integral multiple of 2π

• The centripetal force in the circular orbit is balanced by the electrostatic force of attraction of the positively charged nucleus
• By using the force balance we can obtain the relationship of radius with orbit number and the atomic number of atom

  The formula for radius by substitution of constant gives

Radius = 0.529×$n^{2}$/z armstrong

where n is the orbit number and the z is the atomic number=1

Calculating

Radius difference=n₂-n₁

⇒0.529×($206^{2}$-$4^{2}$)

=0.529×20=10.58 Armstrong