MHTCET 16 May 2024 Shift 1 Paper

Concept:The de-Broglie wavelength of an electron in a Bohr orbit is directly proportional to the principal quantum number $n$.Explanation:According to Bohr's model, the circumference of the $n^{\text{th}}$ orbit equals an integral multiple of the de-Broglie wavelength:$$2 \pi r_n = n \lambda$$Here, $r_n$ is the radius of the $n^{\text{th}}$ orbit, and $\lambda$ is the de-Broglie wavelength.Rearranging the equation gives:$$\lambda = \frac{2 \pi r_n}{n}$$For a hydrogen atom, the orbital radius depends on the principal quantum number as:$$r_n \propto n^2$$Substituting this relation into the wavelength expression:$$\lambda \propto \frac{n^2}{n} = n$$Thus, the de-Broglie wavelength increases as $n$ increases.When the electron jumps from the ground state ($n = 1$) to a higher excited state ($n = 2, 3, \dots$), the value of $n$ rises.So, the de-Broglie wavelength associated with the electron will increase.Answer:Option D: will increase.