Question

An electron change its Position from orbit \(n=4\) to the orbit \(\mathrm{n}=2\) of an atom the wave length of emitted radiation in the form of \(\mathrm{R}\) (where \(\mathrm{R}\) is Redburg constants) (A) \((16 / 7 \mathrm{R})\) (B) \((16 / \mathrm{R})\) (C) (16 / 3R) (D) \((16 / 5 \mathrm{R})\)

Short answer

The wavelength of the emitted radiation when an electron changes its position from orbit n=4 to orbit n=2 is given by the formula \[\lambda = \frac{16}{3R}\]. Therefore, the correct answer is (C) 16 / 3R.

Step-by-step solution

Identify the given variables

In this problem, we are given the initial and final orbits of the electron: n₁ = 4 (Initial orbit) n₂ = 2 (Final orbit) The Rydberg constant for hydrogen is approximately R = 1.097 x 10^7 m⁻¹.

Plug in the values into the Rydberg formula

Now, we will plug in the values of n₁ and n₂ into the Rydberg formula: \[\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = R \left( \frac{1}{4^2} - \frac{1}{2^2} \right)\]

Simplify the equation

Now we will simplify the equation: \[\frac{1}{\lambda} = R \left( \frac{1}{16} - \frac{1}{4} \right)\] Next, find the common denominator (16) and subtract the fractions: \[\frac{1}{\lambda} = R \left( \frac{1-4}{16} \right)\] And finally, simplify the expression: \[\frac{1}{\lambda} = R \left( -\frac{3}{16} \right)\]

Solve for λ in terms of R

Now we need to solve for λ in terms of R. First, multiply both sides by -16/3: \[\lambda = -\frac{16}{3R}\] Since the wavelength cannot be negative, we can take the absolute value: \[\lambda = \frac{16}{3R}\] The wavelength of the emitted radiation is (C) 16 / 3R.