The third line of the Balmer series in the emission spectrum of the hydrogen atom is due to the transition from the

Question

The third line of the Balmer series in the emission spectrum of the hydrogen atom is due to the transition from the (A) fourth Bohr orbit to the first Bohr orbit (B) fifth Bohr orbit to the second Bohr orbit (C) sixth Bohr orbit to the third Bohr orbit (D) seventh Bohr orbit to the third Bohr orbit

Tardigrade Admin · Accepted Answer

In Balmer series, n1=2 and n2 = 3, 4, 5 So, third line arises by the emission of 5th orbit to 2nd orbit.

Q. The third line of the Balmer series in the emission spectrum of the hydrogen atom is due to the transition from the

fourth Bohr orbit to the first Bohr orbit

fifth Bohr orbit to the second Bohr orbit

sixth Bohr orbit to the third Bohr orbit

seventh Bohr orbit to the third Bohr orbit

In Balmer series, $n_{1} = 2$ and $n_{2} = 3$ , $4$ , $5$ So, third line arises by the emission of $5^{t h}$ orbit to $2^{n d}$ orbit.

Q. The third line of the Balmer series in the emission spectrum of the hydrogen atom is due to the transition from the

fourth Bohr orbit to the first Bohr orbit

fifth Bohr orbit to the second Bohr orbit

sixth Bohr orbit to the third Bohr orbit

seventh Bohr orbit to the third Bohr orbit

In Balmer series, n1​=2 and n2​=3, 4, 5 So, third line arises by the emission of 5th orbit to 2nd orbit.

In Balmer series, $n_{1} = 2$ and $n_{2} = 3$ , $4$ , $5$ So, third line arises by the emission of $5^{t h}$ orbit to $2^{n d}$ orbit.