Advertisements
Advertisements
प्रश्न
The energy levels of an atom are as shown below. Which of them will result in the transition of a photon of wavelength 275 nm?

Advertisements
उत्तर
Energy transitions for A,B,C, and D are:
A = 2 eV
B = 4.5 eV
C = 2.5 eV
D = 8 eV
`E=(hC)/lamda`
Where,
E = Energy transition
λ = Wavelength
h = 6.63 × 10−34 Js
C = 3 × 108 m/s
For B, we have
`lambda = (6.63 xx 10^-34 xx 3 xx 10^8)/(4.5 xx 1.6 xx 10^-19)`
`lambda = 275`nm
Thus, B will result in transition of a photon of wavelength of 275 nm.
APPEARS IN
संबंधित प्रश्न
The total energy of an electron in the first excited state of the hydrogen atom is about −3.4 eV.
What is the potential energy of the electron in this state?
The total energy of an electron in the first excited state of the hydrogen atom is about −3.4 eV.
Which of the answers above would change if the choice of the zero of potential energy is changed?
What are means by pair annihilation? Write a balanced equation for the same.
Wavelengths of the first lines of the Lyman series, Paschen series and Balmer series, in hydrogen spectrum are denoted by `lambda_L, lambda_P and lambda_B` respectively. Arrange these wavelengths in increasing order.
A 12.3 eV electron beam is used to bombard gaseous hydrogen at room temperature. Upto which energy level the hydrogen atoms would be excited?
Calculate the wavelengths of the second member of Lyman series and second member of Balmer series.
Draw the energy level diagram showing how the line spectra corresponding to Paschen series occur due to transition between energy levels.
Which transition corresponds to emission of radiation of maximum wavelength?
The Ionisation energy of hydrogen atom is 3.6 ev The ionisation energy of helium atom would be
Energy of an electron at infinity from nucleus is ______.
The diagram shows the four energy levels of an electron in the Bohr model of the hydrogen atom. Identify the transition in which the emitted photon will have the highest energy.
