InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 301. |
The angular momenta of the electron in successive Bohr orbits differ by(A) \(\cfrac{nh}{2\pi}\)(B) h (C) \(\cfrac h{2\pi}\)(D) (n – 1) \(\cfrac h{2\pi}\) |
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Answer» Correct option is (C) \(\cfrac h{2\pi}\) |
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| 302. |
How does the spectrum emitted by rarefied gases different from those dense gases? |
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Answer» In the rarefied gases ,the separation between atoms or molecules are farther apart. Hence the atoms give discrete wavelengths without any interaction with the neighbouring atoms. |
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| 303. |
Give an account of the spectral series of hydrogen atom. |
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Answer» Hydrogen atom has a single electron. Its spectrum consists of series of spectral lines. Lyman series : Lyman Series consists of spectral lines corresponding to the transition of an electron from higher energy orbits n = 1 and n2 = 2,3,4… These lines belong to Ultraviolet region. Balmer series : Balmer series consists of spectral lines emitted during transitions of electrons from higher energy orbits to the second orbit. n1 = 2 and n2 = 3,4,5… .These lines lie is the visible region. Paschen series : Paschen series consists of spectral lines emitted when electron jumps higher energy orbits to the third orbit n1 = 3 and n2 = 4,5,6 … .These lines lie in the infrared region. Brackett series : Brackett series consists of spectral lines emitted during transitions of electrons from higher energy orbits to fourth orbit. n1 = 4 and n2 = 5,6,7… Pfund series : Pfund series consists spectral lines emitted during transition of electrons from higher energy orbits to the fifth orbit. . n1 = 5 and n2 = 6,7,8… . These lines lie in infrared region . The transition from (n1 + 1) to n1 corresponding to Ist member or longest wavelength of the series. The transition from (infinity) state to ‘n1’ state corresponds to the last number or series limit or shortest wavelength of the series. |
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| 304. |
if three are N atoms in a source of laser light and each atom is emitting light with intensity l, then the total intensity produced by it isA. NIB. `N^2I`C. `N^3I`D. `N^4I` |
| Answer» Correct Answer - B | |
| 305. |
What are the characteristics of nuclear forces ? |
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Answer» The three basic characteristics of nuclear forces are as follows: 1. Nuclear forces are very strong attractive forces. 2. Nuclear forces are short range forces. 3. Heavy nuclei for their stability must also have a repulsive component such that the nuclear forces are on the whole attractive. |
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| 306. |
What is radioactivity? OR Define radioactivity. |
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Answer» Radioactivity is the phenomenon in which unstable nuclei of an element spontaneously distintegrate into nuclei of another element by emitting a or β particles accompanied by γrays. Such transformation is known as radioactive transformation or radioactive decay. |
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| 307. |
State the nature and properties of a-particles. |
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Answer» Nature of α-particles : (1) An alpha particle is a helium nucleus, i.e., a doubly ionized helium atom. It consists of two protons and two neutrons. (2) Mass of the α-particle ≅ 4u Charge on the α-particle = 2 × charge on the proton Properties of α-particles : (1) Of the three types of radioactive radiations, α- particles have the maximum ionizing power. It is about 100 times that of β-particles and 104 times that of γ-rays. (2) They have the least penetrating power, about 100 times less than that of β-particles and 104 times less than that of γ-rays. They can pass through very thin sheets of paper but are scattered by metal foils and mica. Since αparticles produce intense ionization in a medium, they lose their kinetic energy quickly. As a result, they do not penetrate more than a few centimetres (about 2.7 cm to 8.6 cm) in air under normal conditions. (3) They are deflected by electric and magnetic fields since they are charged particles. Their deflection is less than that of β-particles in the same field. (4) They affect photographic plates. (5) They cause fluorescence in fluorescent materials such as zinc sulphide. (6) They emerge from the nuclei with tremendous speeds in the range of 1/100th to 1/10th of the speed of light in free space. (7) They destroy living cells. [Note : α-rays and β-rays were discovered by Henri BecquereL] |
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| 308. |
What is the shortest wavelength present in the Paschen series of spectral lines? |
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Answer» Rydberg’s formula is given as \(\frac{hc}{\lambda} = 21.76 \times 10^{-19}(\frac{1}{n_1^2}-\frac{1}{n_2^2})\) h = 6.6 x 10-34 is the plank's constant. c = 3 x 108 is speed of life. The shortest wavelength present in the Paschen series of the spectral lines is given for values n1 = 3 and n2 = \(\infty\) From the above values, \(\lambda = 819nm\) |
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| 309. |
The ionisation energy of hydrogen is 13.6 eV . The energy of the photon released when an electron jumps from the first excited state (n=2) to the ground state of hydrogen atom isA. 3.4 eVB. 4.53 eVC. 10.2 eVD. 13.6eV |
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Answer» Correct Answer - C The energy rquired to removed electron in the first excited state (n=2) The energy of the photon `Rhc((1)/(1^(2))-(1)/(n^(2)))` `rArr" " 13.6((1)/(4)-(1)/(4))=13.6((4-1)/(4))` `" " =13.6 xx(3)/(4)=(40.8)/(4)=10.2eV` |
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| 310. |
Find the radius of the hydrogen atom in its ground state. Also calculate the velocity of the electron in n = 1 orbit. Given `h=6.63xx10^(-34)J s, m=9.1xx10^(-31) kg,` `e=1.6xx10^(-19)C, K=9xx10^(9)Nm^(2)C^(-2)` |
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Answer» `n=1, h=6.63xx10^(-34) J-s,` `m=9.1xx10^(-31)kg` `e=1.6xx10^(-19)C,` `K=9xx10^(9) Nm^(2)C^(-2)` i) `r=(n^(2)h^(2))/(4pi^(2)"mKe"^(2))` `=((1)^(2)xx(6.63xx10^(-34))^(2))/(4xx(3.14)^(2)xx9.1xx10^(-31)xx9xx10^(9)xx(1.6xx10^(-9))^(2))` `therefore r=0.5315 Å` ii) `V=sqrt((KZ)/(mr))xxe` `=sqrt((9xx10^(9)xx1)/(9.1xx10^(-31)xx0.53xx10^(-10)))xx1.6xx10^(-19)` `therefore V=2.18xx10^(6) ms^(-1)` |
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| 311. |
Calculate the ionization energy for a lithium atom. |
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Answer» For `3^(Li^(7))` atom, `Z=3, n=2 [because Li =1s^(2) 2s^(1)]` `E_(n)=(13.6Z^(2))/(n^(2))eV` `=(13.6xx(3)^(2))/(4)=30.6 eV` `therefore` Ionization energy of Lithium =30.6 eV. |
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| 312. |
Prove that the ionisation energy of hydrogen atom is 13.6 eV. |
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Answer» n=1 corresponds to ground state. `E=(-13.6)/(n^(2))eV` `E=(-13.6)/(1^(2))eV` `E=-13.6 eV` `therefore` The minimum energy required to free the electron from the ground state of hydrogen atom =13.6 eV. `therefore` Ionisation energy of hydrogen atom = 13.6 eV. |
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| 313. |
Electrons with de- Broglie wavelength `lambda` fall on the target in an X- rays tube . The cut off wavelength of the emitted X- rays isA. `lambda_(0)=(2mclambda^(2))/(h)`B. `lambda_(0) =(2h)/(mc)`C. `lambda_(0)=(2m^(2)c^(2)lambda^(3))/(h^(2))`D. `lambda_(0)=lambda` |
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Answer» Correct Answer - A Cut-off wavelength occures when incoming electron losses its complete energy is collison . This energy appears in the from X-rays . Given , mass of electrons =m de-Broglie wavelength =`lambda` so, kinetic energy of electron `=(p^(2))/(2m)=((h/(lambda)))/(2m)=(h^(2))/(2mlambda^(2))` Now, maximum energy of photon cab be given by `" " E=(hx)/(lambda_(0))=(h^(2))/(2mlambda^(2))rArrlambda_(0)=(hcxx2lamda^(2)*m)/(h^(2))=(2mclambda^(2))/(h)` |
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| 314. |
The nucleus radius of `._(8)O^(16)` is `3 xx 10^(-15)m`. Find the density of nuclear matter. |
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Answer» Correct Answer - `2.359 xx 10^(17) kgm^(-3)` |
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| 315. |
What is a radioactive element? Give two examples of radioactive elements. |
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Answer» The element which exhibits the property of radioactivity, i.e., spontaneous disintegration of unstable nuclei of the element by emission of α or β particles accompanied by γ-rays is called a radioactive element. Examples : Uranium, thorium, polonium, radium, actinium. |
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| 316. |
Define (i) atomic number (ii) mass number. Give their symbols. |
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Answer» 1. The number of protons in the nucleus of an atom of an element is called the atomic number of the element. It is also known as the proton number. It is denoted by Z. 2. The number of nucleons (protons and neutrons) in the nucleus of an atom is called the mass number or the atomic mass number. It is denoted by A. [Notes : (i) The number of neutrons in the nucleus of an atom is known as the neutron number, denoted by N. It is usually greater than Z, with the exceptions of helium (2 protons, 2 neutrons) and hydrogen (1 proton, no neutron) (ii) A = Z + N.] |
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| 317. |
Write the atomic symbol for an element giving the atomic number and mass number. Give two examples. Which of the two numbers is characteristic of the element? Why? |
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Answer» An atom is represented as \(^A_ZX\), where X is the chemical symbol for the element, Z is the atomic number and A is the mass number. Examples : Fluorine, \(^{19}_9F\); Phosphorus, \(^{31}_{15}p\); Gold, \(^{197}_{73}Au.\) The atomic number of an element, which is the number of protons in the nucleus of an atom of the element, is characteristic of the element. It equals the number of electrons in the atom and hence determines the chemical properties of the element and its place in the modern periodic table. |
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| 318. |
Give de-Broglie’s explanation of quantisation of angular momentum as proposed by Bohr. |
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Answer» The condition for stationary wave formation is that the total distance travelled between the nodes(two) up and down or given path is integral multiple of ‘λ’ i.e ., 2πrn = nλ where, n = 1,2,3,……. But λ = h/mv (from de-Broglie’s hypothesis) 2πrn = nh/mvn mvnrn = n (h/2π) i.e. integral multiple of (h/2π) should be equal to the angular momentum of electron in the orbit. |
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| 319. |
Write the mathematical conditions for quantisation of orbits and energy states. |
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Answer» (i) mvr= nh/2π → quantisation of angular momentum (ii) E2 – E1 = hν → quantisation of energy states and resulting transition. |
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| 320. |
Mention any two demerits of Bohr’s Theory. |
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Answer» (i) The theory is applicable only for hydrogen atom. (ii) The relativistic variation of mass is not taken into account in the theory. (iii) The fine structure of spectral lines cannot be accounted for. (iv) The theory fails to account for relative intensities of spectral lines. |
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| 321. |
Energy evolved from the fusion reaction `2_(1)^(2)H=_(1)^(4)He+Q` is to be used for the production of power. Assuming the efficiency of the process to be `30%`. Find the mass of deuterium that will be consumed in a second for an output of `50 MW`. |
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Answer» Correct Answer - 25.417 g |
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| 322. |
Name two particles found inside the nucleus. What are they collectively called? |
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Answer» Protons and Neutrons. Mass number. |
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| 323. |
Give the charge and the mass of an electron, proton and neutron present in an atom. |
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| 324. |
The ratio of the largest to shortest wavelength in Balmer series of hydrogen spectra is,A. `(25)/(9)`B. `(17)/(6)`C. `(9)/(5)`D. `(5)/(4)` |
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Answer» Correct Answer - C `" " lambdaprop(1)/((1/(n_(f^(2)))-1/n_(i^(2))))` `" " (lambda_("longest"))/(lambda_("shortest"))=(((1)/(4)-(1)/(oo)))/((1/(4)-(1)/(9)))=((1//4))/((5//36))=(9)/(5)` |
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| 325. |
The electron in hydrogen atom makes a transition `n_(1)ton_(2)` where `n_1` and `n_2` are the principal quantum number of two states. Assuming the Bohr model to be valid, the time period of the electron in the initial state is eight times that in the final state. The possible value of `n_1` and `n_2` are:A. `n_1=4,n_2=2`B. `n_1=8,n_2=2`C. `n_1=8,n_2=1`D. `n_1=6,n_2=2` |
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Answer» Correct Answer - A In the `n^(th)` orbit , let `r_n=` radius and `v_n=` speed of electron. Time period , `T_n=(2pir_n)/(v_n)prop (r_n)/(v_n)` Now, `r_npropn^2` and `v_nprop(1)/(n)therefore (r_n)/(v_n)propn^3` or `T_npropn^3`. Here, `8=((n_1)/(n_2))^3` or `(n_1)/(n_2)=2 ` or `n_1-2n_2` |
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| 326. |
How many spectral series are possible in the hydrogen spectrum? |
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Answer» Infinite. The last (sixth) of the named series in the hydrogen spectrum is the Humphreys series which results from transitions to nf = 6 from ni = 7, 8, 9, … etc. in the far-infrared region. Further series are unnamed but follow the same pattern. Their lines are increasingly faint showing that they correspond to increasingly rare atomic events. [Note : There are infinite number of lines in each series. The spacing between adjacent lines decreases with decreasing wavelength, converging to the so-called series limit or short-wavelength limit.] |
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| 327. |
A hydrogen atom and a `Li^(2+)` ion are both in the second excited state. If `l_H` and `l_(Li)` are their respective electronic angular momenta, and `E_H and E_(Li)` their respective energies, then(a) `l_H gt l_(Li) and |E_H| gt |E_(Li)|`(b) `l_H = l_(Li) and |E_H| lt |E_(Li)|` (C ) `l_H = l_(Li) and |E_H| gt |E_(Li)|`(d) `l_H lt l_(Li) and |E_H| lt|E_(Li)|`A. `l_(H)gtl_(Li)` and `|E_(H)|gt|E_(Li)|`B. `l_(H)=l_(Li)` and `|E_(H)|gt|E_(Li)|`C. `l_(H)=l_(Li)` and `|E_(H)|lt|E_(Li)|`D. `l_(H)lt l_(Li)` and `|E_(H)|lt|E_(Li)|` |
| Answer» Correct Answer - C | |
| 328. |
In hydrogen atom, the Balmer series is obtained when the electron jumps from (A) a higher orbit to the first orbit (B) the first orbit to a higher orbit (C) a higher orbit to the second orbit (D) the second orbit to a higher orbit. |
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Answer» (C) a higher orbit to the second orbit |
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| 329. |
In Rutherford scattering experiment, what will b ethe correct angle for `alpha` scattering for an impact parameter `b = 0` ? |
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Answer» We know that impact parameter, b`propcot.(theta)/(2)rArr` `b=K cot. (theta)/(2)` where, K is a constant Here, if, b=0 `" " Kcot theta//2=0rArrcot theta//2=0rArr(theta)/(2)=(pi)/(2)` Hence, `" " theta=pi=180^(@)` |
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| 330. |
(A) atoms of each element are stable and emit characteristic spectrum. (R) the spectrum provides useful information about the atomic structure.A. if both assertion and reason are true and reason is the correct explanation of assertion.B. if both assertion and reason are true but reason is not the correct explanation of assertion.C. if assertion is true but reason is false.D. if both assertion and reason are false. |
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Answer» Correct Answer - B Atoms of each element are stable and emit characteristic spectrum. The spectrum consists of a set of isolated parallel lines termed as line spectrum . It provides useful information about the atomic structure . |
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| 331. |
(A) atom as a whole is electrically neutral. (R)atom contains equal amount of positive and negative charges.A. if both assertion and reason are true and reason is the correct explanation of assertion.B. if both assertion and reason are true but reason is not the correct explanation of assertion.C. if assertion is true but reason is false.D. if both assertion and reason are false. |
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Answer» Correct Answer - A For an atom, number of electrons is equal to number of protons and neutrons are electrically neutral , so net charge on an atom is nill. |
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| 332. |
At what speed must an electron revolve around the nucleus of hydrogen atom so that it may not be pulled into the nucleus by electrostatic attraction? Given, mass of electron `=9.1xx10^(-31) kg`, radius of orbit `=0.5xx10^(-10)m and e=1.6xx10^(-19)C.` |
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Answer» Correct Answer - `2.25 xx 10^(6) ms^(-1)` |
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| 333. |
Beta rays emitted by a radioactive material are (1) electromagnetic radiations (2) the electrons orbiting around the nucleus(3) charged particles emitted by the nucleus (4) neutral particle |
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Answer» Correct option : (3) charged particles emitted by the nucleus Explanation : From the basic studies of the phenomenon of radioactivity, it is known that β particles are fast moving electrons emitted by the nucleus. One can confirm this by determining the e/m values of these particles. |
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| 334. |
The half -life of polonium is 138 days against alpha decay . What fraction of a sample of pure polonium will remain undecayed after 276 days ? |
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Answer» Correct Answer - `(1)/(4)` |
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| 335. |
The half life of Radon is 3.8 days. Calculate how much of 15 milligram of Radon will remain after 38 days. |
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Answer» Correct Answer - 0.015 mg |
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| 336. |
A set of atom in an excited state decaysA. in general to any of the states with lower energyB. into a lower state only when excited by an external electric fieldC. all togather simultaneously into a lower stateD. to emit photons only when they collide |
| Answer» Correct Answer - A | |
| 337. |
Let `F_1` be the frequency of second line of Lyman series and `F_2` be the frequency of first line of Balmer series then frequency of first line of Lyman series is given byA. `F_1-F_2`B. `F_1+F_2`C. `F_2-F_1`D. `(F_1F_2)/(F_1+F_2)` |
| Answer» Correct Answer - A | |
| 338. |
the wavelength of the first line of lyman series for hydrogen atom is equal to that of the second line of balmer series for a hydrogen like ion. The atomic number Z of hydrogen like ion isA. 4B. 1C. 2D. 3 |
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Answer» Correct Answer - C Lyman series of H-atom we, can write `(hc)/(lambda)=Rhc((1)/(1^(2))-(1)/(2^(2)))` where, symbols hav their usual meaning. And for second line of Balmer series f H-like ion `" " (hc)/(lambda)=Z^(2)Rhc((1)/(2^(2))-(1)/(4^(2)))` Therefore ` ((1)/(2^(2))-(1)/(2^(2)))=Z^(2)((1)/(4)-(1)/(16))` `" " (1-(1)/(4))=Z^(2)((1)/(4)-(1)/(16))rArrZ=2.19=2` |
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| 339. |
Find the ratio of wavelengths of first line of Lyman series and second line of Balmer series |
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Answer» Use Rydberg formula `1/(lambdak_alpha)=R_H(1/1^2-1/2^2)` `1/(lambda_(Lbeta))=R_H(1/2^2-1/4^2)` `(lambda_(kalpha))/(lambda_(Lbeta))=((16-4)xx4xx1)/(4xx16(4-1))` `=12/(16xx3)=1/4` |
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| 340. |
The lines in Balmer series have their wavelengths lying betweenA. 1266 Å to 3647 ÅB. 642 Å to 3000 ÅC. 3647 Å to 6563 ÅD. Zero to infinity |
| Answer» Correct Answer - C | |
| 341. |
Identify the incorrect relationshipA. Number of waves in an orbit, `n=(2pir)/lambda`B. Number of revolutions of an electron per second in `n^"th"` orbit =`V_n/(2pir_n)`C. Wavelength of an electron =`h/p`D. Speed of a (de-Broglie wavelength ) particle accelerated by potential difference V is `v="2eV"/m` |
| Answer» Correct Answer - D | |
| 342. |
If `He^+` ion undergoes transition n=2 `to` 1 the ratio of final to initial magnetic field due to motion of electron at the nucleus willA. `32:1`B. `1:32`C. `16:1`D. `1:16` |
| Answer» Correct Answer - A | |
| 343. |
The electron in a hydrogen atom makes a transition from M shell to L-shell. The ratio of magnitude of initial to final acceleration of the electron isA. `9:4`B. `81:16`C. `4:9`D. `16:81` |
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Answer» Correct Answer - D `a=omega^(2)r` As `" " omegaprop(1)/(n^(3))" " and " : r propn^(2)` `" "aprop(1)/(n^(4))` `rArr" " (a_(M))/(a_(L))=((2)/(3))^(4)=(16)/(81)` |
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| 344. |
The magntic moment `(mu)` of a revolving electron around the mucleaus varies with principle quantum number `n` asA. `mupropn`B. `muprop1//n`C. `mupropn^(2)`D. `muprop1//n^(2)` |
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Answer» Correct Answer - A Magnetic moment `mu` of a revolving electron is given by `mu=(e)/(2m)L=(e)/(2m).(nh)/(2pi)=((eh)/(2m)).(n)/(2pi)=(n)/(2pi)mu_(B)` Hence `" " mu_(B)=(eh)/(2m)` is Bohr magneton `:. " " mu propn` |
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| 345. |
The electron in a hydrogen atom makes a transition from `n=n_(1)` to `n=n_(2)` state. The time period of the electron in the initial state `(n_(1))` is eigh times that in the final state `(n_(2))`. The possible values of `n_(1)` and `n_(2)` areA. `n_(1)=8,n_(2)=1`B. `n_(1)=4, n_(2)=2`C. `n_(1)=2,n_(2)=4`D. `n_(1)1, n_(2)=8` |
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Answer» Correct Answer - B In a hydrogen atom the time period is given by `Tpropn^(3)` `(T_(1))/(T_(2))=((n_(1))/(n_(2)))^(3)rArr(8)/(1) ((n_(1))/(n_(3)))^(3)rArr(n_1)/(n_(2))=(2)/(1)` ltbnrgt Thus, the values must be `n_(1)=4 and n_(2)=2` |
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| 346. |
An electron revolves round a nucleus of charge Ze. In order to excite the electron from the `n=20` to `n=3`, the energy required is `47.2 eV`. Z is equal toA. 3B. 4C. 5D. 2 |
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Answer» Correct Answer - C `Z^(2)[(13.6)/(4)-(13.6)/(9)]=47.2 ` ` :. " " Z=5` |
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| 347. |
In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelangth in the Balmer series is:A. `(5)/(27)`B. `(3)/(23)`C. `(7)/(29)`D. `(9)/(31)` |
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Answer» Correct Answer - A `(1)/(lambda)=R[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` For longest wavelelngth model states that an atom is a sphere of Lyman series `n_(1)=1 and n_(2)=2` For longest wavelength of Balmer series `n_(1)2 and n_(2)=3` `" " So (lambda_("Lyman"))/(lambda_("Balmer"))=(5)/(27)` |
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| 348. |
If the series limit wavelength of the Lyman series for hydrogen atom is `912 Å`, then the series limit wavelength for the Balmer series for the hydrogen atom is |
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Answer» For series limit of Balmer series, we can write, `n_(2)=2,n_(1)=oo` `rArr " " (1)/(lambda)=R((1)/(n_(2)^(2))-(1)/(n_(1)^(2)))=R[(1)/((2)^(2))-(1)/((oo)^(2))]=(R)/(4)` `:. " " lambda=(4)/(R) =(4)/(10967800)=4xx912xx10^(-10)m` `=4xx912Å=3648Å` |
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| 349. |
If the wavelength of the first member of Balmer series in hydrogen spectrum is 6563 Å, calculate the wavelength of the first member of Lymen series in the same spectrum. |
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Answer» Correct Answer - 1215.14 Å |
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| 350. |
The wavelengths of some of the spectral lines obtained in hydrogen spectrum are `1216 Å, 6463 Å and 9546 Å`. Which one of these wavelengths belongs to the Paschen series ? |
| Answer» The wavelength of spectral line `9546 Å` belongs to the Paschen series. | |