An electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom. Given the Rydberg's constant R = 105 cm-1.The freqency in H2 of the emitted radiation will be:

Explanation

The explanation is given below: To find the frequency of the emitted radiation when an electron jumps from the 4th orbit to the 2nd orbit of a hydrogen atom, you can use the Rydberg formula: ν = R_H * (1/n_1^2 - 1/n_2^2), where: ν is the frequency of the emitted radiation. R_H is the Rydberg constant (given as 10^5 cm^(-1)). n_1 is the principal quantum number of the initial orbit (4th orbit). n_2 is the principal quantum number of the final orbit (2nd orbit). Substitute the values into the formula: ν = 10^5 cm^(-1) * (1/4^2 - 1/2^2) Calculate the values inside the parentheses: ν = 10^5 cm^(-1) * (1/16 - 1/4) Now, calculate the values inside the parentheses: ν = 10^5 cm^(-1) * (0.0625 - 0.25) ν = 10^5 cm^(-1) * (-0.1875) Now, calculate the frequency: ν ≈ -1.875 x 10^4 Hz The emitted radiation will have a frequency of approximately -1.875 x 10^4 Hz. The negative sign indicates that the emission is in the form of absorption rather than emission. This means the electron has absorbed energy to jump from the 4th to the 2nd orbit, as such a transition would not naturally occur in a hydrogen atom. Please check if the orbit numbers are correctly provided in the question, as typical transitions involve electrons moving to higher orbits when absorbing energy and to lower orbits when emitting energy.