Question

for the transition from n =2 to n=1 which of the following will produce shortest wavelength?



1. H atom 2. D atom 3.He+ atom 4.Li 2+ atom



explain with clarity,

Answer 1

he+ atom
becos u can put formula:
E= R (1/n1^2-1/n2^2)......or see in ncert 2 chapter
ryberg's constant formula u will get ur ans. its a soloved example too there in ncert read it✌

Answer 2

Answer: The correct answer is option 4.

Explanation: The Rydberg's equation for the calculation of wavelength is given by:

$\frac{1}{\lambda}=R_HZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

where,

$\lambda$ = Wavelength

R = Rydberg's Constant = $109677cm^{-1}$

Z = Atomic Number

$n_1$ = lower energy level

$n_2$ = Upper energy level

As from the equation, the atomic number and wavelength are inversely related. More the atomic number, the shortest will be the wavelength.

From the given options above, the atomic number of lithium is 3 and is the highest of all. So, this nucleus will have the shortest wavelength.