Atomic StructureHard
Question
The ratio of time taken by electron in revolution around the H-nucleus in the second and third orbits is
Options
A.2: 3
B.4: 8
C.8: 27
D.27: 8
Solution
$T_{n,z} = 1.5 \times 10^{- 16} \times \frac{n^{3}}{z^{2}}\text{ sec}$
$\therefore\frac{T_{2,M}}{T_{3,M}} = \frac{2^{3}}{3^{3}} = \frac{8}{27}$
Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
In a measurement of the quantum efficiency of photosynthesis in green plants, it was found that 9 quanta of red light at...If an electron has spin quantum number of + ½ and magnetic quantum number of –1, then it cannot be present in...Which of the following ions has the maximum number of unpaired d-electrons?...When an electron jumps from nth orbit to 1st orbit in an imaginary atom obeying Bohr’s model, it emits two radiations of...In H-atom, if ′x′ is the radius of the first Bohr orbit, de Broglie wavelength of an electron in 3rd orbit i...