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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.
| 401. |
Why line spectra is regarded as the fingerprints of atoms ? |
| Answer» The line spectra of atom of every element is unique. Therefore line spectra which helps in studying the electronic structure of atom results in the identification of the element by matching the spectra of unknown element with that of the known element as is done in the case of fingerprinting. | |
| 402. |
The wave function of 2s electron is given by `Psi_(2s) = (1)/(4 sqrt(2pi)) ((1)/(a_(0)))^(3//2) (2 - (r)/(a_(0))) e^(-r//a_(0))` It has a node at `r = r_(0)`. Find the relation between `r_(0) and a_(0)` |
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Answer» The probability of finding 2s electron will be : `Psi_(2s)^(2) = (1)/(32pi) ((1)/(a_(0)))^(3) (2- (r_(0))/(a_(0)))^(2) e^(-2r//a_(0))` Node is the point at which probability of finding electron is zero. Thus, `Psi_(2s)^(2) = 0` when `r = r_(0)` `:. (1)/(32pi) ((1)/(a_(0)))^(2) (2 - (r_(0))/(a_(0)))^(2) e^(-2r_(0)//a_(0)) = 0` In this expression, the only factor that can be zero is `(2 - (r_(0))/(a_(0)))` Thus, `2 - (r_(0))/(a_(0)) = 0 or r_(0) = 2 a_(0)` |
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| 403. |
Assertion. In Lyman series of H-spectra, the maximum wavelength of lines is 121.56 nm Reason. Wavelength is maximum when the transition is from the very next levelA. If both assertion and reason are true, and reason is the true explanation of the arrertionB. If both assertion and reason are true, but reason is not the true explanation of the assertionalC. If assertion is true, but reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - A `(1)/(lamda) = R_(H) Z^(2) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` For Lyman series, `n_(1) = 1`. For the next level, `n_(2) = 2`. Hence, `(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))` will be minimum or `lamda` will be maximum. `((1)/(lamda) = 109678 xx (1)^(2) [(1)/(1^(2)) - (1)/(2^(2))] cm^(-1) = 109678 xx (3)/(4) cm^(-1) or lamda = (4)/(3) xx (1)/(109678) cm = 121.56 xx 10^(-7) cm = 121.56 xx 10^(-9) m = 121.56 nm)` |
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| 404. |
Thershold frequency, `v_(0)` is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency `1.0xx10^(15)s^(-1)` was allowed to hit a metal surface, an electron having `1.988x10^(-19)J` of kinetic energy was emitted. Calculated the threshold frequency of this metal. equal to 600nm hits the metal surface. |
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Answer» We know that `hv=hv_(0)+KE` `hv-KE=hv_(0)=(6.626xx10^(-34)Jsxx1xx10^(15)s^(-1))-1998xx10^(-19)J` ltbr. `hv_(0)=6.626xx10^(-19)-1.988xx10^(-19)J` `hv_(0)=4.638xx10^(-19)J` `v_(0)=(4.638xx10^(-19)J)/(6.626xx10^(-34)Js)=0.699xx10^(15s^(-1)` when `lambda=600nm=600xx10^(-19)m` `v=(c)/(lambda)=(30xx10^(8)ms^(-1))/(6.0xx10^(-7)m)=0.5xx10^(15)s^(-1)` Thus `vltv_(0)` hence , no electron will be emitted. |
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| 405. |
STATEMENT -1 `:` Work function of a metal depends on ionisation energy. STATEMENT-2` :` One photon can eject one electron STATEMENT -3 `:` K.E. of ejected depends on intensity of lightA. T F FB. T T FC. T F TD. T T T |
| Answer» Correct Answer - B | |
| 406. |
In H-atoms, the energy of electron in the nth orbit is given as `E_(n) = - (13.6)/(n^(2)) eV` Show that `E_((n +1)) - E_(n) = (13.6 xx 2)/(n^(3)) eV` for large value of n. |
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Answer» `E_((n +1)) - E_(n) = - (13.6)/((n +1)^(2)) - (-(13.6)/(n^(2))) = 13.6 [(1)/(n^(2)) - (1)/((n +1)^(2))] = 13.6 xx ((n +1)^(2)- n^(2))/(n^(2) (n +1)^(2))` `= (13.6 xx (2n +1))/(n^(2) (n +1)^(2)) = (13.6 xx 2n)/(n^(2) (n +1)^(2)) + (13.6)/(n(n + 1)^(2)) = (13.6 xx 2)/(n(n +1)^(2)) + (13.6)/(n^(2) (n + 1)^(2))` As n is large, `n + 1 ~=n`. Hence, `E_((n +1)) - E_(n) ~= (13.6 xx 2)/(n xx n^(2)) + (13.6)/(n^(2) xx n^(2)) = (13.6 xx 2)/(n^(3))` (Neglecting second term because `n^(4)` is very large) |
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| 407. |
How many peacks are present in the radial distribution function for the `3d` orbital?A. ZeroB. OneC. TwoD. Three |
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Answer» Correct Answer - B The number of peaks in the radial distribution function is always equal to the number of radial nodes `+1`. For `nd` orbitals, the number of radial nodes is equal to `n - 3`. |
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| 408. |
`ns` orbital has________radial nodes.A. `n`B. `n -3`C. `n - 2`D. `n - 1` |
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Answer» Correct Answer - D For any atomic orbital, the number of radial or spherical nodes is always equal to `n -l -1`. Since `l = 0`, for an `s` orbital, `ns `orbital has `n - 1` radial nodes. |
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| 409. |
A physicist was performing experiments to study the effect of varying voltage on the velocity and wavelength of the electrons. In one case, the electron was accelerated through a potential difference of 1kV and in the second case, it was accelerated through a potential difference of 2kV The velocity acquired by the electron will beA. double in the second case than in the first caseB. four times in the second case than in the first caseC. same in both casesD. 1.4 times in the second case than in the first case |
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Answer» Correct Answer - D K.E. of the electron in case `I = 1000 eV` `= 1000 xx 1.602 xx 10^(-16) J = 1.602 xx 10^(-16)J` K.E. of the electron in case II = 2000 eV `= 2 xx 1.602 xx 10^(-16) J` `:. ((1)/(2) mv_(2)^(2))//((1)/(2) mv_(1)^(2)) = 2 or (v_(2)//v_(1))^(2) = 2` or `v_(2)//v_(1) = sqrt2 = 1.4` |
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| 410. |
A physicist was performing experiments to study the effect of varying voltage on the velocity and wavelength of the electrons. In one case, the electron was accelerated through a potential difference of 1kV and in the second case, it was accelerated through a potential difference of 2kV The wavelength associated with the electron will beA. double in the second case than in the first caseB. double in the first case than in the second caseC. 1.4 times in the second case than in the first caseD. 1.4 times in the first case than in the second case |
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Answer» Correct Answer - D `lamda = (h)/(mv) :. (lamda_(1))/(lamda_(2)) = (v_(2))/(v_(1)) = 1.4, " i.e., " lamda_(1) = 1.4 lamda_(2)` |
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| 411. |
Balmer series of the hydrogen spectrum lies in the ........region |
| Answer» Correct Answer - visible | |
| 412. |
The energy possessed by one mole of photons is called........ |
| Answer» Correct Answer - one einstein | |
| 413. |
How many electrons in sulphur (Z = 16) can have `n + l = 3` ? |
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Answer» Electronic configuration of `._(16)S = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(4)` For `1s^(2), n + l = 1 + 0 = 1`. For `2s^(2), n + l = 2 + 0 =2`. For `2p^(6), n + l = 2 + 1 = 3` For `3s^(2), n + l = 3 + 0 = 3` For `3p^(4), n + l = 3 + 1 = 4` Thus, `n + l = 3 " for " 2p^(6) and 3s^(2)` electrons, i.e., for 8 electrons |
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| 414. |
Based on equation `E = -2.178 xx 10^-18 J((Z^2)/(n^2))`, certain conclusions are written. Which of them is not correct ?A. For n = 1, the electron has a more negative energy than it does for n=6 which means that the electron is more loosely bound in the smallest allowed orbit.B. The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus.C. Larger the value of n, the larger is the orbit radius.D. Equation can be used to calculate the change in energy when the electron changes orbit. |
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Answer» Correct Answer - A The electron is more firmly bound in the smallest allowed orbit. |
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| 415. |
Based on equation `E = -2.178 xx 10^-18 J((Z^2)/(n^2))`, certain conclusions are written. Which of them is not correct ?A. The large the value of `n`, the large is the orbit radius.B. The eqaution can be used to calculate the change in enegry when the electrons change orbit.C. For `n = 1`, the electron has more negative energy tham it does for `n = 6` which means that the electron is more lossely bound in the smallest allowed orbit.D. The negative sign in equation means that the enegry of the electron bound to the nucleus is lower than it would be if the electrons were at infinite distance form the nucleus. |
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Answer» Correct Answer - C More nagative enegry implies that the electron is closer to the nucleus and thus, more strongly bound to the nucleus. `(1)` Size of the orbit is directly related to the value of principal quantum number. (2)` DeltaE = E_(f) - E_(i)` `=- 2.1789 xx 10^(-18)JZ^(2) ((1)/(n_(f)^(2))-(1)/(n_(f)^(2)))` `(4)` At infinite distance form the nucleus, the enegry of electronb is taken to be zero. |
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| 416. |
STATEMENT-1`:` de-Broglie wavelength is not appreciable for molecule`//` atom moving in gaseous state. STATEMENT -2 `:` Law of uncertainty is not applicable to molecule`//`atom of a gas STATEMENT-3 `:` Value of spin quantum number is the directional cosine of magnetic moment dueto rotation of electronA. T T FB. F T TC. T F TD. F F T |
| Answer» Correct Answer - D | |
| 417. |
If the uncertainty in the position of a moving electron is equal to its de Broglie wavelength, then its velocity will be completely uncertain. Explain. |
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Answer» According to available information , `Deltax=lambda` But ` lambda =(h)/(mv) =(h)/(p) :. P=(h)/(lambda)=(h)/(Deltax) or Delta x =(h)/(p)` According to uncertainty principle, `DeltaxDeltap ge (h)/(4pi)`. `:. " " (h)/(p) Deltap ge (h)/(4 pi) or (Deltap)/(p) ge (1)/(4 pi)` (constant) This means that `Deltap ge p or m DeltaV ge mV or DeltaV ge V`. This means that the velocity of electron will be completely uncertain. |
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| 418. |
The number of electrons which will together weigh one gram isA. `1.098xx10^(27)` electronsB. `9.1096xx10^(31)` electronsC. 1 electronD. `1xx10^(4)` electrons |
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Answer» Correct Answer - A Mass of an electron = `9.1096xx10^(-31)kg 1gor10^(-3)kg=1/(9.1096xx10^(-31))xx10^(-3) =1.098xx10^(27)` electrons |
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| 419. |
To which orbit the electron in the hydrogen atom will jump on absorbing `12.1 eV` of energy? |
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Answer» Energy of electron in the nth orbit of H-atom is `E_(n) = (-13.6)/(n^(2)) eV :. E_(1) = -13.6 eV` After absorbing 12.1 eV, the energy will be `= -13.6 + 12.1 eV = -1.5 eV` Thus, `(-13.6)/(n^(2)) = - 1.5 eV or n^(2) = 9 or n = 3` Thus, electron will jump to 3rd orbit. |
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| 420. |
Calculate the wavelength, frequency and wave number of a light wave whose period is `2.0 xx 10^(-10)s` |
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Answer» Frequency `(v) =(1)/("Period") = (1)/(2.0xx 10^(-10)s) = 5 xx 10^(9) s^(-1)` Wavelength, `lamda = (c)/(v) = (3.0 xx 10^(8) ms^(-1))/(5 xx 10^(9) s^(-1)) = 6.0 xx 10^(-2) m` Wave number, `bar(v) = (1)/(lamda) = (1)/(6 xx 10^(-2)m) = 16.66 m^(-1)` |
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| 421. |
What is the ratio between the energies of two radiations, one with a wavelength of `6000 Å` and the other with `2000 Å [1 Å = 10^(-10)m]`? |
| Answer» Correct Answer - `(1)/(3)` | |
| 422. |
Which of the following has largest de-Broglie wavelength, provided all have equal velocityA. Carbon dioxide moleculeB. ElectronC. Ammonia moleculeD. Proton |
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Answer» Correct Answer - B `lamda = h//m v, " i.e., " lamda prop 1//m` |
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| 423. |
Calculate the frequency, wave number and energy associated with photon of radiations having wavelength 6000 Ã… . |
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Answer» Wavelength `(lambda)=6000 Ã…= 6000 xx 10^(-10)m=6 xx 10^(-7)m` Frequency (v) `=(c)/(lambda) =((3xx 10^(8)ms^(-1)))/((6xx 10^(-7)m))=5 xx 10^(14)s^(-1)` Wave number `(bar(v))=(1)/(lambda)=(1)/((6xx10^(-7)m))=1.6 xx 10^(6)m^(-1)` Energy (E) `=hv=(6.626 xx 10^(-34)Js)xx (5xx10^(14)s^(-1))=33.1 xx 10^(-20)J`. |
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| 424. |
When moving with the same velocity which of the following particles has the largest de Broglie wavelength and why ? (i) Electron (ii) Proton (iii) `alpha`-particle. |
| Answer» Electron with least mass has maximum wavelength `(lambda=h//m upsilon)`. | |
| 425. |
What is the ratio between the energies of two radiations one with a wavelength of 6000Ã… and other with 2000Ã…? |
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Answer» Correct Answer - `E_(2)=3E_(1)` `lambda_(1)6000 Ã… , lambda_(2)=2000Ã…`. ltbegt `E_(1)=h (C)/(lambda_(1)) "and" E_(2)=(hc)/(lambda_(2))` `(E_(1))/(E_(2))=(lambda_(2))/(lambda_(1))=(2000)/(6000)=(1)/(3) " " E_(2)=3E_(1)` |
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| 426. |
Calculate the wavelength of a photon in Angstrons having an energy of 1 electron-volt. |
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Answer» Energy of photon ` 1eV=1.602 xx 10^(-19)J` `h=6.626 xx 10^(-34)Js, c=3xx10^(8)ms^(-1)` `E=hv=(hc)/(lambda)` or `lambda= (hc)/(E)` By substituting the values, `" " lambda=((6.626 xx 10^(-34) Js)xx (3xx10^(8)ms^(-1)))/((1.602 xx 10^(-19)J))=12.42 xx 10^(-7)m` `=12.42 xx 10^(-7) xx 10^(10) Ã… or 12.42 xx 10^(3) Ã…`. |
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| 427. |
Schrodinger wave equation for a particle in a one-dimension box isA. `(delta^(2) psi)/(deltax^(2)) + (2m)/(h) (E - oo) psi = 0`B. `(delta^(2)psi)/(deltax^(2)) + (8pi^(2)m)/(h^(2)) (E - V) psi = 0`C. `(delta^(2) psi)/(delta x^(2)) + (2m)/(h) (E -V) psi = 0`D. `(delta^(2) psi)/(delta x^(2)) + (2pi^(2) m)/(h^(2)) (E - oo) = 0` |
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Answer» Correct Answer - B Schrodinger wave equation for a particel in one dimension box is `(delta^(2) Psi)/(delta x^(2)) + (8pi^(2) m)/(h^(2)) (E - V) Psi = 0` |
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| 428. |
Orientation of orbitals is given byA. Magnetic quantum numberB. Spin quantum numberC. Azimuthal quantum numberD. Principal quantum number |
| Answer» Correct Answer - 1 | |
| 429. |
What would be the IUPAC named for element with atomic number 120 ?A. UnunbiumB. UnnibiumC. UnniluniumD. Unbinilium |
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Answer» Correct Answer - D 120 = un + bi + nil + ium = unbinilium |
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| 430. |
Why electronic energy is negative ? Comment on the spacing between the energy levels. |
| Answer» The spacing between the energy levels decreases as we move outwards from the nucleus. | |
| 431. |
The state `S_(1)` isA. 1sB. 2sC. 2pD. 3s |
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Answer» Correct Answer - B It is given that the state `S_(1)` has one radial node, i.e., No. of radial nodes = (n - l - 1) = 1. It is possible only when state `S_(1)` is 2s with n = 2 and l = 0 ( since `S_(1)` is sphericlly symmetrical). |
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| 432. |
How many protons are present in 5.6 L of oxygen at N.T.P.using O-16 isotope only ? |
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Answer» No. of `O_(2)` molecules in 5.6 L of oxygen `=((6.022xx10^(10^(23))xx(5.6L))/((22.4L)))=1.506xx10^(23)` No. of oxygen atoms present = `1.506xx10^(23)xx2` No. of protons in each oxygen atom =8 Total no. of protons present `=2xx1.506xx10^(23)xx8=2.41xx10^(24)`. |
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| 433. |
An element with mass number `81` contains `31.7%` more neutrons as compared to protons. Write the symbol for the isotope in the standard fromat. Strategy: Let us assume that the number of protons `=p`. Then, Number of neutrons `(n) = p +31.7%`of `p` |
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Answer» Mass number `=` Number of protons `+` Number of neutrons `81=p +n` or `p+(p+(31.7)/(100)xxp)=81` `p+(p +0.317p) = 81` `p+p(1+0.317) = 81` `p+1.317p = 81` `(1+1.317)p = 81` `2.317 p = 81` `p = (81)/(2.317) 34.95~= 35` According to the periodic table, the element with atomic number `35` is bromine, Br. Therefore, the symbol is `_(35)^(81)Br`. |
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| 434. |
A cricket ball weighing 100 g is located within 1 nm. What is the uncertainty in the velocity ? |
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Answer» Correct Answer - `5.27xx10^(-25)m s^(-1)` According to uncertainty principle `Delta x *m Deltav=(h)/(4 pi) or Deltav=(h)/(4pixxmxxDeltax)` `h=6.626xx10^(-34)"kg"m^(2)s^(-1), m=100g=0.1 "kg", Deltax=1nm=10^(-9)m, pi=3.143` `Deltav=[((6.626xx10^(-34)"kg" m^(2)s^(-1)))/(4xx3.143xxx(0.1"kg")(10^(-9)m))]" " =5.27xx10^(-25)m s^(-1)`. |
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| 435. |
What values of magnetic quantum number, m are permitted for an electron having angular momentum quantum number value, `l = 2`? |
| Answer» Correct Answer - `-2, -1, 0, +1, +2` | |
| 436. |
What are the value of n, l and m for `2p_(x) and 3p_(z)` orbitals ? |
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Answer» For `2p_(x), n = 2, l = 1, m = +1` For `3p_(z), n = 3, l = 1, m = 0` |
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| 437. |
The numbers of electrons, protons, and neutrons in a species are equal to `18,16`, and `16`, respectively. Write the symol for the species in the standard format. Strategy: Before usding the standard notation `_(Z)^(A)X`, find out wheter the species is a neutral atom, a cation, or an anion. if it is a neutral atom, Eq is valid, i.e. Number of protons = Number of electrons = Atomic number If the species is an ion, determine whether the number of protons is larger (cation, a positive ion) or smaller (anion, a negative ion) than the number of electrons. Number of neurtons is always given by Eq. i.e., `N=A-Z`, whether the species is neutral or charged. |
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Answer» Number of protons `=` Atomic number `= 16` The element, according to the periodic table, is sulphur `(S)`. Number of protons `+` Number of neutrons `=` Mass number `= 16 + 16 = 32` As the number of protons is not equal to the number of electrons, the species is not neutral, but is negatively charged (anion) due to two extra electorns. Thus, the symbol is `._(16)^(32)S^(2-)` |
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| 438. |
The number of electrons,protons, neutron in a species are `18,16 `and `16` respectively. Assigns proper symbols |
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Answer» the atomic number is equal to number of protons = 16. the element is sulphur (s). Atomic mass number = number of protons + number of neutrons = 16 + 16 = 32 species is not neutral as the number of protons is not equal to electrons . It is anion (negatively charged ) with charge equal to excess electrons = 18 - 16 =2 . symbol is `._(16)^(32)S^(2-)` note : before using the notation `._(Z)^(A)X` ,find out whether the species is a netral atom. a cation or an anion. if it is a neutral atom , equatoion (2.3) is valid , i.e number of protons = number of electrons = atoms number . if the species is an ion , determine whether the number of protons are larger (cartion , postive ion ) or smaller ( anion , negative ion) than the number of elctrons . number of neutrons is always given by A-Z , whether the species is neutral or ion. |
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| 439. |
The correct graph regarding v vs KE. ( Incident)A. B. C. D. |
| Answer» Correct Answer - 1 | |
| 440. |
A ball of mass 100 g is moving with a velocity of 100 `msec^(-1)`. Find its wavelength. |
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Answer» According to de Broglie equation, `lambda =(h)/(mv)` `m=100g =0.1 kg , v= 100 ms^(-1), h= 6.626 xx 10^(-34) kg m^(2)s^(-1)` `lambda =((6.626 xx 10^(-34) kg m^(2)s^(-1)))/((0.1 kg )xx(100 ms^(-1)))=6.626 xx 10^(-35)m`. |
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| 441. |
How many electrons can be filled in all the orbitals with `n+l=5` ? |
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Answer» (n+l)=5 has 5s, 4p and 3d orbitals with two, six and ten electrons respectively. Therefore, the total number of electrons = 18 |
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| 442. |
Answer the following : (a) How many electrons can be filled in all the orbitals with n + l= 5 ? (b) Which of the two is paramagnetic, V (IV) orV (V) and why ? (c) How many unpaired electrons are present in Pd (Z=46)? (d) The ion of an element has configuration `[Ar] 3d^(4)` in +3 oxidation state. What will be the electronic configuration of its atom ? |
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Answer» (a) (n+1)=5 has 5s, 4p and 3d orbitals with two, six and ten electrons respectively.Therefore, the total number of electrons is 18. (b) The configuration are , V (IV) , `[Ar]^(18)3d^(1)` and `V (IV) : [Ar]^(18)`. Therefore , V (IV) is paramagnetic in nature due to the presence of unpaired electron. (c) The electronic configuration of the element palladium `(Z=46) "is" [Kr]^(36)4d^(10)5s^(0)`. Therefore , it has no unpaired electron. The configuration of ion is `[Ar]3d^(4)`. Therefore , the configuration of the atom is `[Ar]3d^(5)4s^(2)`. |
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| 443. |
Total number of orbitals associated with third shell will be...A. 2B. 4C. 9D. 3 |
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Answer» Correct Answer - C No. of orbitals in 3rd shell `(n =3) = n^(2) = 3^(2) = 9` |
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| 444. |
How many electrons in an atom have the following auantum numbers ? (i) `n=4, m_(s)=-1/2` (ii) `n=3, l=0`. |
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Answer» (i) Total electrons in `n=4` are `2 n^(2)`, i.e., `2xx4^(2)=32`. Half of them, i.e., 16 electrons have `m_(s)=-1/2`. (ii) `n=3, l=0` means 3s orbital which can have 2 electrons. |
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| 445. |
If n = 5, how many electrons can have `m_(l) = +1`? |
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Answer» When n = 5, l = 0, 1, 2 , 3 and 4 When l = 0, `m != +1` When l = 1, 2, 3, or 4, in each case, there is one orbital which has `m_(l) = +1` `(l = 1, m = -1, 0 +1, l = 2, m = -2, -1, 0 +1, +2`, etc) Hence, there are four orbitals which will have `m_(l) = +1` EAch orbital can have maximum two electrons. Hence, electrons with n = 5 and `m_(l) + 1` will be 8 |
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| 446. |
Calculate the kinetic energy of the electron ejected when yellow light of frequency `5.2 xx 10^(14) sec^(-1)` falls on the surface of potassium metal. Threshold frequency of potassium is `5 xx 10^(14) sec^(-1)` |
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Answer» K.E. of the ejected electron is given by `(1)/(2) mv^(2) = hv - hv_(0) = h (v - v_(0))` `= 6.625 xx 10^(-34) Js (5.2 xx 10^(14) - 5.0 xx 10^(14)) s^(-1) = 6.625 xx 10^(-34) xx 0.2 xx 10^(14)` joules `= 1.325 xx 10^(-20)` Joules |
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| 447. |
what should be the velocity of an electron so that its momentum becomes equal to that of a photon of wavelength `5200 Å` |
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Answer» According to de Broglie equation, `lambda =(h)/(m upsilon) ` Momentum of electron , ` m upsilon = (h)/(lambda)=((6.626 xx 10^(-34)kg m ^(2) s^(-1)))/((5200 xx 10^(-10)m))=1.274 xx 10^(-27) kg ms^(-1) ...(i)` The momentum of electron can also be calculated as `= m upsilon = (9.1 xx 10^(-31)kg) xx upsilon " " ...(ii)` Comparing (i) and (ii) `(9.1 xx 10^(-31)kg)xx upsilon = (1.274 xx 10^(-27) kg ms^(-1))` ` :. " " upsilon= ((1.274 xx 10^(-27) kg ms^(-1)))/((9.1 xx 10^(-31) kg))= 1.4 xx 10^(3) m s^(-1)` |
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| 448. |
Calculate the momentum of a particle which has a de Brglie wavelength of `1 Å` or 0.1 nm. `(h = 6.6 xx 10^(-34) kg m^(2) s^(-1))` |
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Answer» Correct Answer - `6.6 xx 10^(-24) kg ms^(-1)` `lamda = (h)/(p) or p = (h)/(lamda) = (6.6 xx 10^(-34) kg m^(2) s^(-1))/(10^(-10)m) = 6.6 xx 10^(-24) kg m s^(-1) (1 Å = 10^(-10)m, 1nm = 10^(-9) m)` |
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| 449. |
Calculate the ratio of the time periods `(T_(1)//T_(2))` in second orbit of hydrogen atom to third orbit of `He^(+)` ion |
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Answer» Time period `(T_(n)) = alpha (n^(2))/(Z^(2))` `:. (T_(1))/(T_(2)) = (n_(1)^(3))/(Z_(1)^(2)) xx (Z_(2)^(2))/(n_(2)^(3)) = (2^(3))/(1^(2)) xx (2^(2))/(3^(3)) = (32)/(27)` |
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| 450. |
Given that for `He^(+)` ion, the difference between the longest wavelength line of Balmer series and Lymen series is 133.8 nm, calculate the value of Rydberg constant. |
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Answer» For H-like particle. `bar(v) = (1)/(lamda) = RZ^(2) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` where R is Rydberg constant For `He^(+), Z = 2` and for longest wavelength line of Balmer series, `n_(1) = 2, n_(2) =3`. Hence, `(1)/(lamda_("Balmer")) = R (2)^(2) ((1)/(2^(2)) - (1)/(3^(2))) = 4R ((1)/(4) - (1)/(9)) = 4R xx (5)/(36) = (5R)/(9) or lamda_("Balmer") = (9)/(5R)` For longest wavelength in Lyman series, `n_(1) =1, n_(2) = 3=2`. Hence, `lamda_("Lyman") = R (2)^(2) ((1)/(1^(2)) - (1)/(2^(2))) = 4R (1 - (1)/(4)) = 4R xx (3)/(4) = 3R or lamda_("Lyman") = (1)/(3R)` Given `lamda_("Balmer") - lamda_("Lyman") = 133.8 xx 10^(-9)m` `:. (9)/(5R) - (1)/(3R) = 133.8 xx 10^(-9) or (1)/(R) ((9)/(5) - (1)/(3)) = 133.8 xx 10^(-9)` or `(1)/(R) xx (22)/(15) = 133.8 xx 10^(-9) or R = (22)/(15) xx (1)/(133.8 xx 10^(-9)) = 1.0961 xx 10^(9) m^(-1)` |
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