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De-broglie's Explanation Of Bohr's Second Postulate - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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Two identical non-relativistic particles more right angles to each other, possessing DeBroglie wavelengths \lambda_1 and \lambda_2 The De-Broglie wavelength of each particle in the frame of their centre of mass.

An electron of mass' m ' and charge ' e ' initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t, ignoring relativistic effects, is-

The de-Broglie wave lengths of a proton and \alpha-particle are equal. The ratio of their velocities are-

The potential energy of a particle of mass m is given by V(x)=\left\{\begin{array}{ll} \epsilon_0, & 0 \leq x \leq 1 \\ 0, & x>1 \end{array}, \lambda_1 \text { and} \ \lambda_2\right. are de-Broglie wavelength of the particle, when \mathrm{0 \leq x \leq 1}\mathrm{0 \leq x \leq 1} and x>1 respectively. If the total energy of the particle is \mathrm{2 E_0,} then the ratio \mathrm{\frac{\lambda_1}{\lambda_2}} will be -

Radiation from hydrogen gas excited to first excited state is used for illuminating certain photoelectric plate. When the radiation from some unknown hydrogen like gas excited to the same level is used to expose the same plate, it is found that the de–Broglie wavelength of the fastest photoelectron has decreased 2.3 times. It is given that the energy corresponding to the longest wavelength of the Lyman series of the unknown gas is 3 times the ionization energy of hydrogen gas (13.6 eV). Find the work function of photoelectric plate in eV. (Take (2.3)^2=5.25

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An electron and a photon have same energy 4 E. The ratio of de -Broglie wavelength of electron to wavelchgeh of photon (Mass of electron is m and speed of light is c)

Electrons with de-Broglie a fall on the target in an x-ray tube. The cut off wave length of the emitted x-rays.
 

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De-broglie's explanation of Bohr's second postulate

De-broglie's explanation of Bohr's second postulate-

Since the Bohr gave many postulates in his theory, but the second postulate is not very clear and little puzzling. The Scientist De Broglie explained this puzzle very clearly that why the angular momentum of the revolving electron is the integral multiple of the h/2 \pi. De broglie in his experiment proved that the electron revolving the circular orbit has wave nature also in the last chapter we have seen the experiment performed by the Davison and Germer which proved that the electron shows the wave nature. In analogy to waves travelling on a string, particle waves too can lead to standing waves under resonant conditions. During the chapter Waves and Oscillation, we know that when a string is plucked, a vast number of wavelengths are excited. However only those wavelengths survive which have nodes at the ends and form the standing wave in the string. It means that in a string, standing waves are formed when the total distance travelled by a wave down the string and back is any integral number of wavelengths. Waves with other wavelengths interfere with themselves upon reflection and their amplitudes quickly drop to zero.

For an electron moving in n^{\text {th }} circular orbit of radius r_{n}, the total distance is the circumference of the orbit, 2 \pi r_{n}.

                                                                          2 \pi r_{n}=n \lambda, \quad n=1,2,3 \dots

                                                                            

Figure given above illustrates a standing particle wave on a circular orbit for n = 4, i.e., 2πrn = 4λ, where λ is the de Broglie wavelength of the electron moving in nth orbit. From the last chapter we have studied that λ = h/p, where p is the magnitude of the electron’s momentum. If the speed of the electron is much less than the speed of light, the momentum is mvn.

                                                                                      Thus,

                                                                                                \lambda = \frac{h}{mv_n}.

From the above equation, we have,

                                                                        2 \pi r_n = \frac{nh}{mv_n} \ \ \ or, \ \ mv_n r_n = \frac{nh}{2 \pi}

This is the quantum condition proposed by Bohr for the angular momentum of the electron. Thus de Broglie hypothesis provided an explanation for Bohr’s second postulate for the quantisation of angular momentum of the orbiting electron.

The quantised electron orbits and energy states are due to the wave nature of the electron and only resonant standing waves can persist. Bohr’s model, involving classical trajectory picture (planet-like electron orbiting the nucleus), correctly predicts the gross features of the hydrogenic atoms(Hydrogenic atoms are the atoms consisting of a nucleus with positive charge +Ze and a single electron, where Z is the proton number. Examples are hydrogen atom, singly ionised helium, doubly ionised lithium, and so forth.), in particular, the frequencies of the radiation emitted or selectively absorbed.

 

 

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De-broglie's explanation of Bohr's second postulate

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