The gravitational force between an H-atom
Question: The gravitational force between an H-atom and another particle of mass m will be given by Newtons law: F = G M.m/r2, where r is in km and (a) M = mproton + m electron (b) M = mproton + melectron B/c2(B = 13.6 eV) (c) M is not related to the mass of the hydrogen atom (d) M = mproton + m electron |V |/c2 (|V | = magnitude of the potential energy of electron in the H-atom) Solution: (b) M = mproton + melectron B/c2(B = 13.6 eV)...
Read More →Suppose we consider a large number of containers
Question: Suppose we consider a large number of containers each containing initially 10000 atoms of a radioactive material with a half-life of 1 year. After 1 year (a) all the containers will have 5000 atoms of the material (b) all the containers will contain the same number of atoms of the material but that number will only be approximately 5000 (c) the containers will, in general, have different numbers of the atoms of the material but their average will be close to 5000 (d) none of the contai...
Read More →The odds in favor of the occurrence of an event are 8 : 13.
Question: The odds in favor of the occurrence of an event are 8 : 13. Find the probability that the event will occur. Solution: We know that, If odds in favor of the occurrence an event are a:b, then the probability of an event to occur is $\frac{a}{a+b}$ which indirectly came from Probability of the occurrence of an event $=\frac{\text { Total no. of Desired outcomes }}{\text { Total no. of outcomes }}$ Where, Total no.of desired outcomes = a, and total no.of outcomes = a+b Given $a=8, b=13$ Th...
Read More →If 7/10 is the probability of occurrence of an event
Question: If 7/10 is the probability of occurrence of an event, what is the probability that it does not occur? Solution: We know that, Probability of occurring = 1 - the probability of not occurring Given the probability of occurrence $=\frac{7}{10}$ Therefore, the probability of not occurrence $=1-\frac{7}{10}$ $=\frac{3}{10}$ Conclusion: Probability of not occurrence is $\frac{3}{10}$...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{x}{\sqrt{8+x-x^{2}}} d x$ Solution: Given $I=\int \frac{x}{\sqrt{-x^{2}+x+8}} d x$ Integral is of form $\int \frac{p x+q}{\sqrt{a x^{2}+b x+c}} d x$ Writing numerator as $\mathrm{px}+\mathrm{q}=\lambda\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right\}+\mu$ $\Rightarrow p x+q=\lambda(2 a x+b)+\mu$ $\Rightarrow x=\lambda(-2 x+1)+\mu$ $\therefore \lambda=-1 / 2$ and $\mu=-1 / 2$ Let $x=-1 / 2(-2 x+...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{3 x+1}{\sqrt{5-2 x-x^{2}}} d x$ Solution: Given $I=\int \frac{3 x+1}{\sqrt{-x^{2}-2 x+5}} d x$ Integral is of form $\int \frac{p x+q}{\sqrt{a x^{2}+b x+c}} d x$ Writing numerator as $\mathrm{px}+\mathrm{q}=\lambda\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right\}+\mu$ $\Rightarrow p x+q=\lambda(2 a x+b)+\mu$ $\Rightarrow 3 x+1=\lambda(-2 x-2)+\mu$ $\therefore \lambda=-3 / 2$ and $\mu=-2$ Let $3 ...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{6 x-5}{\sqrt{3 x^{2}-5 x+1}} d x$ Solution: Given $I=\int \frac{6 x-5}{\sqrt{3 x^{2}-5 x+1}} d x$ Integral is of form $\int \frac{\mathrm{px}+\mathrm{q}}{\sqrt{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}}} \mathrm{dx}$ Writing numerator as $\mathrm{px}+\mathrm{q}=\lambda\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right\}+\mu$ $\Rightarrow p x+q=\lambda(2 a x+b)+\mu$ $\Rightarrow 6 x-5=\lambda(6 x-5)+\...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{x+1}{\sqrt{x+5 x-x^{2}}} d x$ Solution: Given $I=\int \frac{x+1}{\sqrt{4+5 x-x^{2}}} d x$ Integral is of form $\int \frac{p x+q}{\sqrt{a x^{2}+b x+c}} d x$ Writing numerator as $p x+q=\lambda\left\{\frac{d}{d x}\left(a x^{2}+b x+c\right)\right\}+\mu$ $\Rightarrow p x+q=\lambda(2 a x+b)+\mu$ $\Rightarrow x+1=\lambda(-2 x+5)+\mu$ $\therefore \lambda=-1 / 2$ and $\mu=7 / 2$ Let $x+1=-1 / 2(-2 x+5)+7 / 2$ $\Rightarrow \int \frac{x+1}{\sqrt{-x^{...
Read More →The Bohr model for the H-atom relies on
Question: The Bohr model for the H-atom relies on the Coulombs law of electrostatics. Coulombs law has not directly been verified for very short distances of the order of angstroms. Supposing Coulombs law between two opposite charge + q1, q2 is modified to $|F|=\frac{q_{1} q_{2}}{\left(4 \pi \epsilon_{0}\right)} \frac{1}{r^{2}}, r \geq R_{0}$ $=\frac{q_{1} q_{2}}{\left(4 \pi \epsilon_{0}\right)} \frac{1}{R_{0}^{2}}\left(\frac{R_{0}}{r}\right)^{\epsilon}, r \leq R_{0} \quad$ Calculate in such a c...
Read More →The inverse square law in electrostatics is
Question: The inverse square law in electrostatics is $|F|=\frac{e^{2}}{\left(4 \pi \epsilon_{0}\right)} r^{2}$ for the force between an electron and a proton. The $1 \mathrm{r}$ dependence of $|\mathrm{F}|$ can be understood in quantum theory as being due to the fact that the particle of light (photon) is massless. If photons had a mass mp, force would be modified to $|F|=\frac{e^{2}}{\left(4 \pi \epsilon_{0}\right)} r^{2}\left[\frac{1}{r^{2}}+\frac{\lambda}{r}\right]$ where $\lambda=\mathrm{m}...
Read More →In the Auger process, an atom makes a transition
Question: In the Auger process, an atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (This is called an Auger electron). Assuming the nucleus to be massive, calculate the kinetic energy of an n = 4 Auger electron emitted by Chromium by absorbing the energy from an = 2 to n = 1 transition. Solution: The energy in the nth state = -Rch Z2/n2= -13.6 Z2/n2eV Where R is the Rydberg constant and Z ...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{2 x+1}{\sqrt{x^{2}+2 x-1}} d x$ Solution: Given $I=\int \frac{2 x+1}{\sqrt{x^{2}+2 x-1}} d x$ Integral is of form $\int \frac{p x+q}{\sqrt{a x^{2}+b x+c}} d x$ Writing numerator as $\mathrm{px}+\mathrm{q}=\lambda\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right\}+\mu$ $\Rightarrow p x+q=\lambda(2 a x+b)+\mu$ $\Rightarrow 2 x+1=\lambda(2 x+2)+\mu$ $\therefore \lambda=1$ and $\mu=-1$ Let $2 x+1=2 x...
Read More →If a proton had a radius R and the charge was uniformly
Question: If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) R = 0.1Å, and (ii) R = 10 Å. Solution: The kinetic energy of the H atom is 13.6 eV Potential energy of an electron and proton is -27.2 eV Total energy of the electron = -13.6 eV When R = 0.1Å, the ground state energy is same as the earlier = -13.6 eV When R = 10 Å, The kinetic energy = 0.16 eV The potential energy of electron and proton = -3.83 ...
Read More →Deuterium was discovered in 1932 by Harold Urey
Question: Deuterium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in 1H and 2H. This is because the wavelength of transition depends to a certain extent on the nuclear mass. If nuclear motion is taken into account then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass , revolving around the nucleus at a distance equal to the electron-nucleus sep...
Read More →Which of the following cannot be the probability of occurrence of an event?
Question: Which of the following cannot be the probability of occurrence of an event? (i) 0 (ii) $\frac{-3}{4}$ (iii) $\frac{3}{4}$ (iv) $\frac{4}{3}$ Solution: (ii) and (iv) cant be the probability of occurrence of an event. So, (ii) and (iv) are the answers to our question. Explanation: We know that $0 \leq$ probability $\leq 1$ i.e. probability can vary from 0 to 1 (both are inclusive) So, (i) 0 can be possible as $0 \leq$ probability $\leq 1$ (ii) $-\frac{3}{4}$ 4 is not possible as it is le...
Read More →The first four spectral lines in the Lyman series
Question: The first four spectral lines in the Lyman series of an H-atom are = 1218 Å, 1028Å, 974.3 Å and 951.4Å. If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines. Solution: Let H and D be the reduced masses of electrons of hydrogen and deuterium respectively. ni and nf are the fixed mass series for hydrogen and deuterium Rh/Rd = H/ D Reduced mass of hydrogen = H = me/1+me/M = me (1-me/M) Reduced mass of deuterium = D = 2M.me/2M(1+me/2M) = me(1-...
Read More →What is the minimum energy that must be given to an H
Question: What is the minimum energy that must be given to an H atom in ground state so that it can emit an H line in Balmer series? If the angular momentum of the system is conserved, what would be the angular momentum of such H photon? Solution: Energy required for the transition from n = 1 to n = 5 is given as: E = E5 E1 = (-13.6/52) (-13.6/12) = 13.06 eV Angular momentum is given as = change in angular momentum of electron = 5(h/2) 2(h/2 ) = 3h/2 = 3.17 10-34Js....
Read More →Show that the first few frequencies of light
Question: Show that the first few frequencies of light that are emitted when electrons fall to the nth level from levels higher than n are approximate harmonics (i.e. in the ratio 1 : 2: 3) when n 1. Solution: Frequency of any line in the series on the spectrum of hydrogen is expressed as the difference between the two atoms fcm = cRZ2[1/(n+p)2 1/n2] where m = n + p Substituting for p we get, fcm = (2cRZ2/n3)p...
Read More →Using the Bohr model,
Question: Using the Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state. Solution: Let v be the velocity of the electron Number of revolutions per unit time = f = 2a0/v The electric current is given as I = Q When change Q flows in time T Such that Q = e Therefore, the electric current is given as I = 2a0/v e...
Read More →Assume that there is no repulsive force between
Question: Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulombs law as usual. Under such circumstances, calculate the ground state energy of a He-atom. Solution: For He nucleus, Z = 2 and in-ground state n = 1 Therefore, En = -13.6 Z2/n2eV = -54.4 eV...
Read More →Positronium is just like an H-atom
Question: Positronium is just like an H-atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium? Solution: The lowest energy of the positronium is -6.8 electron volts The highest energy level of the positronium which is the next to n = 1 that is n = 2 is -1.7 electron volt....
Read More →Consider two different hydrogen atoms.
Question: Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model? Solution: As the value of n in two different hydrogen atoms are different, their angular momentum will be different. According to Bohrs model, the angular momentum is given as L = nh/2...
Read More →Would the Bohr formula for the H-atom remain unchanged
Question: Would the Bohr formula for the H-atom remain unchanged if proton had a charge (+4/3)e and electron a charge (3/4)e, where e = 1.6 1019C? Give reasons for your answer. Solution: As there is no change in the position of the proton and electron the product of both the charges will remain the same....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{x}{\sqrt{x^{2}+6 x+10}} d x$ Solution: Given $I=\int \frac{x}{\sqrt{x^{2}+6 x+10}} d x$ Integral is of form $\int \frac{\mathrm{px}+\mathrm{q}}{\sqrt{\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}}} \mathrm{dx}$ Writing numerator as $\mathrm{px}+\mathrm{q}=\lambda\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right\}+\mu$ $\Rightarrow \mathrm{px}+\mathrm{q}=\lambda(2 \mathrm{ax}+\mathrm{b})+\mu$ $\Rightarro...
Read More →When an electron falls from a higher energy
Question: When an electron falls from a higher energy to a lower energy level, the difference in the energies appears in the form of electromagnetic radiation. Why cannot it be emitted as other forms of energy? Solution: When an electron falls from a higher energy to a lower energy level, the difference in the energies appears in the form of electromagnetic radiation because there is an acceleration of the charged particle....
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