PYQSide
ExamsMy practice
The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is
(2) 4
The last line (series limit) of a series corresponds to the transition from ni = .
Balmer series (nf = 2): 1 B = R ( 122 - 1 2 ) = R4
B = 4R
Lyman series (nf = 1): 1 L = R ( 112 - 1 2 ) = R
L = 1R
B L = 4/R1/R = 4
Answer: (2)
Given the value of Rydberg constant is 107 m-1, the wave number of the last line of the Balmer series in hydrogen spectrum will be
When an -particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on m as
Electrons of mass m with de-Broglie wavelength fall on the target in an X-ray tube. The cutoff wavelength ( 0) of the emitted X-ray is
If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength . When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be
In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is
Consider 3rd orbit of He+ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K = 9 109 constant, Z = 2 and h (Planck's Constant) = 6.6 10-34 J s]
Hydrogen atom in ground state is excited by a monochromatic radiation of = 975 A. Number of spectral lines in the resulting spectrum emitted will be
Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is
An electron in hydrogen atom makes a transition n1 n2 where n1 and n2 are principal quantum numbers of the two states. Assuming Bohr's model to be valid, the time period of the electron in the initial state is eight times that in the final state. The possible values of n1 and n2 are
The transition from the state n = 3 to n = 1 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from