MHT CET 2019 3rd May Morning Shift | Atoms and Nuclei Question 89 | Physics

In hydrogen emission spectrum, for any series, the principal quantum number is $n$. Corresponding maximum wavelength $\lambda$ is ( $R=$ Rydberg's constant)

$\frac{R(2 n+1)}{n^2(n+1)}$

$\frac{n^2(n+1)^2}{R(2 n+1)}$

$\frac{n^2(n+1)}{R(2 n+1)}$

$\frac{R(2 n+1)}{n^2(n+1)^2}$

MHT CET 2019 2nd May Evening Shift

MCQ (Single Correct Answer)

-0

When the electron in hydrogen atom jumps from fourth Bohr orbit to second Bohr orbit, one gets the

second line of Balmer series

first line of Balmer series

first line of Pfund series

second line of Paschen series

MHT CET 2019 2nd May Evening Shift

MCQ (Single Correct Answer)

-0

In Balmer series, wavelength of first line is ' $\lambda_1$ ' and in Brackett series wavelength of first line is ' $\lambda_2$ ' then $\frac{\lambda_1}{\lambda_2}$ is

0.162

0.124

0.138

0.188

MHT CET 2019 2nd May Morning Shift

MCQ (Single Correct Answer)

-0

Bohr model is applied to a particle of mass ' $m$ ' and charge ' $q$ ' is moving in a plane under the influence of a transverse magnetic field ' $B$ '. The energy of the charged particle in the $n$th level will be ( $h=$ Planck's constant)

$2 n h q B / \pi \mathrm{m}$

$n h q B / 2 \pi \mathrm{~m}$

$n h q B / 4 \pi \mathrm{~m}$

$n h q B / \pi \mathrm{m}$

PREVIOUS NEXT

MHT CET Subjects