The colour of the second line of the Balmer s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The Balmer series corresponds to transitions where the electron jumps to the $n=2$ orbit.The first line corresponds to the transition from $n=3$ t

Vedclass · Accepted Answer

Blue

Vedclass · Answer

(A) The Balmer series corresponds to transitions where the electron jumps to the $n=2$ orbit.The first line corresponds to the transition from $n=3$ to $n=2$.The second line corresponds to the transition from $n=4$ to $n=2$.The wavelength $\lambda$ is calculated using the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right)$.Calculating this gives $\lambda \approx 486.1 \ nm$.This wavelength falls in the blue-green region of the visible spectrum,which is commonly identified as blue.

The colour of the second line of the Balmer series is

Similar Questions

Which series of the hydrogen spectrum lies in the visible region?

An electron changes its position from orbit $n = 4$ to the orbit $n = 2$ of an atom. The wavelength of the emitted radiation is ($R =$ Rydberg's constant).

The shortest wavelength in the Lyman series of the hydrogen spectrum is $912 \ \mathring{A}$,corresponding to a photon energy of $13.6 \ eV$. The shortest wavelength in the Balmer series is about..... $\mathring{A}$.

If the first line in the Lyman series has wavelength $\lambda$,then the first line in the Balmer series has the wavelength

The wavelength of the second Balmer line in the hydrogen spectrum is $600 \ nm$. The wavelength for its third line in the Lyman series is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module