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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.
| 1901. |
Calculate the radius ratio of `2^(nd)` excited state of `H & 1^(st)` excited state of `Li^(+2)`. |
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Answer» `2^(nd)` excited state, means `e^(-)` is present in `3^(rd)` shell of hydrogen `r_(3) = 0.529xx((3)^(2))/(1)=0.529xx9` `1^(st)` excited state, means `e^(-)` exist in `2^(nd)` shell of `Li^(+2)` `r_(2)=0.529xx((2)^(2))/(3)` `=0.529xx(4)/(3)implies ((r_(3))_(H))/((r_(2))_(Li^(+2)))=(0.529xx(9)/(1))/(0.529xx(4)/(3))` `=("radius of " 2^(nd) " excited state of hydrogen")/("radius of " 1^(st) " excited state of " Li^(+2)) implies ((r_(3))_(H))/((r_(2))_(Li^(+2)))=(27)/(4)` |
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| 1902. |
Find radius (A) `1^(st)` Bohr orbit of H-atom (B) `2^(nd)` shell of `Li^(+2)` ion `(Z=3)` (C ) `M` shell of `He^(+)` ion `(Z=2)` |
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Answer» `r_(n) = 0.529xx(n^(2))/(Z)Å` (A) `r_(1)=0.529xx(1^(2))/(1)=0.529Å` (B) `r_(2)=0.529xx(2^(2))/(3)=(4)/(3)(0.529)Å` (C ) `r_(3)=0.529xx(3^(3))/(2)=(9)/(2)(0.529)Å` |
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| 1903. |
`1^(st)` excitation Potential of a hydrogen like sample is 15 volt. If all the atoms of the sample are in `2^(nd)` excited state then find the `K.E.` in `eV` of the electron ejected if a photon of energy `(65)/(9)eV` is supplied to this sample.A. 5B. 2C. 6D. 4 |
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Answer» Correct Answer - 1 `15eV=13.6Z^(2)xx(3)/(4)" "rArr13.6Z^(2)=20eV` Energy of 2nd excited state `=(13.6Z^(2))/(3^(2))=(-20)/(9)eV, :. K.E.` of electron `=(65)/(9)-(20)/(9)=(45)/(9)=5eV` |
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| 1904. |
`1^(st)` excitation Potential of a hydrogen like sample is 15 volt. If all the atoms of the sample are in `2^(nd)` excited state then find the `K.E.` in `eV` of the electron ejected if a photon of energy `(65)/(9)eV` is supplied to this sample.A. 54.4 eVB. 24 eVC. 122.4 eVD. 216 eV |
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Answer» Correct Answer - D |
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| 1905. |
Find total energy, `PE & KE` of electron in : (A) `2^(nd)` orbit of `He^(+)` ion. (B) `1^(st)` excited state of `Be^(+3)` ion. |
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Answer» `E_(n) =- 13.6 xx(Z^(2))/(n^(2)) (eV)/("atom"),E_(n) =- KE = (PE)/(2)` `(A) E_(2)=-13.6xx(2^(2))/(2^(2))=-13.6" "(B) E_(3)=-13.6xx(4^(2))/(2^(2))=-54.4eV` `KE = 13.6eV, PE = - 27.2 eV" "KE = 54.4 eV, PE=-108.8eV` |
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| 1906. |
A photon of energy `15 eV`collision , H-atom gets ionised. The maximum kinetic energy of emitted electron isA. `1.4 eV`B. `5 eV`C. `15 eV`D. `13.6 eV` |
| Answer» Correct Answer - A | |
| 1907. |
For the energy level is an atom which one of the following statement is correct ?A. The 4s sub-energy level is at a higher energy than the 3d sub-energy level.B. The M-energy level can have maximum of 32 electrons.C. The second principal energy level can have four orbital and contain a maximum of 8 electrons.D. The 5th main energy level can have maximum of 50 electrons. |
| Answer» Correct Answer - C | |
| 1908. |
A new electron enters the orbital when:A. (n+l) is minimumB. (n+l) is maximumC. (n+m) is minimumD. (n+m) is maximum |
| Answer» Correct Answer - A | |
| 1909. |
Among the following paires of orbitals. Which orbital will the lagrest effective anuclear charge ?`2s and 3s` 4d and Delta f ` `3d and 3p` |
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Answer» `2s` , because `2s` orbital lies nearer to the nucleaus than `3s` orbital b. `4d` lies to the nucleus c. `3p` lies mear to the nucleus |
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| 1910. |
For a given value of n (principal quantum number), the energy of different subshells an be arranged in the order of:A. `fgtdgtpgts`B. `sgtpgtdgtf`C. `fgtpgtdgts`D. `sgtfgtpgtd` |
| Answer» Correct Answer - A | |
| 1911. |
When electric transition occurs from higher energy state to lower energy state with energy difference equal to `Delta E` electron volts , the wavelkength of the line emitted is apporomately equal toA. `(12395)/(Delta E ) xx 10^(-10)m`B. `(12395)/(Delta E ) xx 10^(10)m`C. `(12395)/(Delta E ) xx 10^(-10)m`D. `(12395)/(Delta E ) xx 10^(10)m` |
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Answer» Correct Answer - A `lambda = (hc)/(Delta E) = (6.62 xx 10^(-34) J s xx 3 xx 10^(8) m s^(-1))/(E xx 1.602 xx 10^(-19) J)` `= 12395 xx 10^(-10)m` |
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| 1912. |
The fundamental particle which are responsible for leping nucless togather isA. MesonB. AntiprotonC. PositronD. Electron |
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Answer» Correct Answer - A Meson |
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| 1913. |
The atomic number at which filling of a g-orbitals is likely to begin is:A. 121B. 116C. 106D. 124 |
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Answer» Correct Answer - A For g-orbital `l = 4 rarr n = 5` or more |
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| 1914. |
The orientation of an orbital is govermed by the quantum number known as ……….. and is rerresented by the suymbol ……………. |
| Answer» Correct Answer - Magnetic ,m | |
| 1915. |
After filling the 4d-orbitals, an electron will enter in:A. 4pB. 4sC. 5pD. 4f |
| Answer» Correct Answer - C | |
| 1916. |
The sequence of filling atomic orbitals is govermed by …….. Principle |
| Answer» Correct Answer - Aufthau | |
| 1917. |
Which of the following set of quantum number is an impossible arrangement ?A. `n = 3,m = -2,s = +1//2`B. `n = 4,m = 3,s = +1//2`C. `n = 5,m = 2,s = -1//2`D. `n = 3,m = -3,s = -1//2` |
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Answer» Correct Answer - D `n = m` |
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| 1918. |
An electron beam can undergo defraction by crystals .Through what potential should a beam of electrons beb acceleration so that its wavelength becomkes equal to `1.54 Å` |
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Answer» For an electron `(1)/(2) mv^(2) = eV` where v in the accelerating portential `lambda = (h)/("mu")` `:. (1)/(2)m((k)/(m lambda))^(2) = eV` `:. V = (1)/(2) xx (h^(2))/(m lambda^(2)e)`, subatituting all value , we get `= (1 xx (6.626 xx 10^(-34))^(2))/( 2 xx 9.108 xx 10^(-13) xx (1.54 xx 10^(-19))^(2) xx 1.602 xx 10^(-19))` |
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| 1919. |
An electron beam can undergo defraction by crystals .Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to `1.54 Å` |
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Answer» For an electron , `1/2 mu^2 =e.V` and ` lambda = ( h)/( mu)` Thus ` 1.2 m h^2 /( m^2lambda^2) = e. V` or ` V= 1/2 h^2/( mlambda^2 . E)` ` = (1 xx ( 6.62 xx 10^(-34)))/(2 xx 9. 108 xx 10^(-31) xx ( 1.54 xx 10^(-10))^2 xx 1. 602 xx 10^(19))` ` 63.3 ` volt . |
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| 1920. |
The number fo electrons present in ` 1 cm^3 ` of ` Mg` (density of `. _(12)Mg^(24) g cm^(-3))` is:A. `5 . 24 xx 10^(23)`B. ` 4 lambda//3`C. `3 lambda`D. `3 lambda//4` |
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Answer» Correct Answer - A `w_(Mg) 1 cm^3 = 1. 74 g` ` :.` No . Of elecrron `= ( 1. 74 xx 12 xx 6.023 xx 10^(23))/(24)` ` = 5.24 . xx 10^(23)`. |
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| 1921. |
Given that the abundacne of isotopes `.^(54)Fe`, `.^(56)Fe`, and `.^(57)Fe` is 5%, 90% and 5% respectively. The atomic mass of `Fe` isA. `55.85`B. `55.95`C. `55.75`D. `56.05` |
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Answer» Correct Answer - B (b) Average atomic mass of `Fe` =`((54 xx 5)+(56 xx 90)+(57 xx 5))/(100) = 55.95`. |
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| 1922. |
Time taken by an electrons to complete one revolution in the Bohr orbit of the `H` atom isA. `(4pi^2mr^2)/(nh)`B. `(nh)/(4pi^2 mr)`C. `(nh)/(4pi^2 mr^2)`D. `h/(2pimr)` |
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Answer» Correct Answer - A It will take `(4pi^2mr^2)/(nh)` |
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| 1923. |
Find out the number of waves made by a bohr electron is one complete revolation in its third orbit |
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Answer» For an electron to be in a particular energy level of radius r, with de brogle wavelength `lambda` `2pi r = n lambda` (where n is the number of weaves in one revolution `lambda = (h)/(mv)` `rArr n = (2 pi rm v)/(h)` For third orbit `r = 0.53 xx 10^(-10) (n)^(2)m V = 2.165 xx 10^(6) xx ((1)/(n))` `= 0.53 xx 10^(-10) xx (3)^(2)m = 2.165 xx 10^(6) xx (1)/(3)` `= 4.77 xx 10^(-10)m = 0.721 xx 10^(6)` Subsuitating the value of `r,v and h` we get `n = (2 pi rmv)/(h)` `= (2 xx 3.14 xx 4.77 xx 10^(-10) xx 0.721 xx 10^(6) xx 9.1 xx 10^(-31))/(6.626 xx 10^(-34))` |
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| 1924. |
Given that the abundacne of isotopes `.^(54)Fe`, `.^(56)Fe`, and `.^(57)Fe` is 5%, 90% and 5% respectively. The atomic mass of `Fe` isA. `55.85`B. `55.95`C. `55.75`D. `55.05` |
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Answer» Correct Answer - B Atomic mass of iron `= (( 5 xx 54) + (90 xx 56) + (5 xx 57))/(100)` `= 55.95` |
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| 1925. |
A gas of identical H-like atom has some atoms in the lowest (ground) energy level A and some atoms in a particular upper (excited) energy level B and there are no atoms in any other energy level. The atoms of the gas make transition to a higher energy level by aborbing monochromatic light of photon energy `2.7eV`. Subsequently, the atoms emit radiation of only six different photons energies. Some of the emitted photons have energy `2.7eV`. Some have more and some have less than `2.7eV`. (a) Find the principal quantum number of initially excitied level B. (b) Find the ionisation energy for the gas atoms. (c ) Find the maximum and the minimum energies of the emitted photons. |
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Answer» Correct Answer - `(a) n=2,(b)14.4eV, (c )13.5eV,0.7eV` Since we obtain 6 emission lines, it means electron comes from 4th orbit energy emitted is equal to less than and more than `2.7eV`. So it can be like this `E_(4)-E_(2)=2.7eV," "E_(4)-E_(3)lt2.7eV`, `E_(4)-E_(1)gt2.7eV` (a) `n=2` `(E_(4)-E_(2))^("atom")=(E_(4)-E_(2))^(H)xxZ^(2)` `2.7=2.55xxZ^(2)=1.029` (b) `IP=13.6Z^(2)=13.6xx(1.029)^(2)=14.4eV` (c ) Maximum energy emitted `=E_(4)-E_(1)=(E_(4)-E_(1))^(H)xxZ^(2)` `=12.75xx(1.029)^(2)` `13.5eV` Minimum energy emitted `=E_(4)-E_(3)=(E_(4)-E_(3))^(H)xxZ^(2)` `=66xx(1.029)^(2)=0.7eV` |
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| 1926. |
A photosensitve metallic surface has work fuction ` hv_0` If photons of ` 2hv_0` fall on the suface , the electrons come out with a maximum velocity of ` 4xx 10^6 m//s` . If photon energy is increases to ` 5hv_0` the maximum velocity of photoelectrons will be:A. ` 2xx 10^7 m//s`B. `82xx 10^6 m//s`C. ` 2xx 10^76m//s`D. ` 8xx 10^5 m//s` |
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Answer» Correct Answer - B ` 2h-0 = hv_0 + K.E._1` , or `hv_0 = K.E. _1 = 1/2 mu_1^2` Now `5 hv_0= hv_0 + K.E._2` `K. E._2 = 4 hv_0 = 1/2 mu_1=2^2` ` :. U_2^2 /u_1^2 =4` or ` u_2 = 2 u_1` . |
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| 1927. |
For a satellite moving in an orbit aroun the earth the ratio of kinetic enrgy and potential enrgy is :A. ` - 1/2`B. ` 2`C. ` sqrt 2`D. ` 1/(sqrt2` |
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Answer» Correct Answer - A `K.E. = + 1/2 e^2 /r-n and p.E. =- e^2/r_n`. |
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| 1928. |
Which of the following corresponds to one node ?A. B. C. D. |
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Answer» Correct Answer - B It is a fact . |
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| 1929. |
Calculate the Rydberg constant ` R_H` if `He^+` ions are known to have the wavelength difference between the from ( of the longest wavength ) lines fo Balmer and Lyman series equal to ` 133.7 nm`. |
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Answer» Correct Answer - `1.096XX10^(7)m^(-1)` For the first of Blamer & Lymam series res pectively for the atom `(1)/(lambda)=RZ^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]=4R[(1)/(2^(2))-(1)/(3^(2))]=(5R)/(9)` `(1)/(lambda)=4[(1)/(3^(2))-(1)/(4^(2))]=(7R)/(36)` `lambda=lambda_(1)-lambda_(2)=(1)/(R)((36)/(7)-(9)/(5))` `R=1.09xx10^(7)m^(1)` |
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| 1930. |
A hydrogen like atom (atomic number Z) is in a higher excited satte of quantum number n .This excited atom can make a transition to the first excited state by succesively emitting two photon of energies `10.20 eV` and `17.00 eV` .Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting twio photon of energy `4.25 ev` and `5.95 eV` Determine the followings: The value of atomic number (Z) isA. 2B. 4C. 6D. 3 |
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Answer» Correct Answer - D `Z= 3` |
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| 1931. |
A hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number n.This excited atom can make transition to the first excited state by succesively emitting two photons of energies `10.20 eV` and `17.00 eV` respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies `4.25 eV` and ` 5.95 eV` respectively. Determine the values of n and z |
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Answer» ` Delta E = Z^(2) (13.6)[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` The electron makes a transition from state `n_(2) ` is the first excited state `(n_(1) = 2)` releasing `10.2 = 17 = 27.2 eV` energy `27.2 = Z^(2)(13.6)[(1)/(4) - (1)/(n_(2)^(2))]`........(i) When the electron comes to the second excited state `(n_(1) = 3) ` the energy is `4.25 = 5.95 = 10.2 eV` Hence `10.2 = Z^(2)(13.6) ((1)/(9) - (1)/(n_(2)^(2)))`......(ii) ON solving equation (i) and (ii) we get `Z = 3 ` and `n_(2) = 6` |
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| 1932. |
A hydrogen like atom (atomic number `z`) is in a higher excited state of quantum number `n`. This excited atom can make a transition to the first excited state by successively emitting two photons of energies `10.2 eV` and `17.0 eV` respectively. Alternatively the atom from the same excited state can make a transition to the second excited state by successively emitting 2 photons of energy `4.25 eV` and`5.95 eV` respectively. Determine the value of `(n+z)` |
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Answer» Correct Answer - 6; 3 |
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| 1933. |
Which one of the following leads to third line of Balmer spectrum from red end (for hydrogen atom)?A. `2to5`B. `5to2`C. `3to2`D. `4to1` |
| Answer» Correct Answer - B | |
| 1934. |
Find the binding energy of an electron in the ground state of a hydrogen like atom in whose spectrum the third Balmer line is equal to 108.5 mm. |
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Answer» Correct Answer - 54.4 eV |
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| 1935. |
` psi^2` ,( psi) the wave function resperesents the probability of finding electron . Its value depends :A. How much it is inside the nucleusB. How much it is far from the the nuycleusC. How much it is near the nucleusD. Upon the type of orbital |
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Answer» Correct Answer - D For s-orbitals, ` psi^(2)` is maxumum for closer to nucleus . For p-bitals ` psi^(2)` maximum for fa away distance from nucleus . |
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| 1936. |
` psi^2` ,( psi) the wave function resperesents the probability of finding electron . Its value depends :A. Inside the nucleusB. Far from the nucleusC. Near the nucleusD. Upon the type of orbital |
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Answer» Correct Answer - D graphical representation of `psi^(2)` value represent orbital. |
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| 1937. |
The probability density plots of 1 s and 2s orbitals are given in figure. The density of dots in a region represetns the probability density of finding electrons in the region. On the basis of above diagram which of the following statements is incorrect?A. 1s and 2s orbitals are spectrical in shapeB. the probability of finding the electron is maximum near the nucleusC. The probability of finding the electron at a given distance is equal in all directionsD. The probability density of electrons for 2s orbital decreases uniformly as distance from the nucleus increases. |
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Answer» Correct Answer - D The probability density of electrons in 2s orbital first increases then decreases and after that it begins to increases again as distance increases from nucleus |
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| 1938. |
`A` particular radiostation broadcasts at a frequency of `1120` Kilo Hertz another radio station broadcasts at a frequency of `98.7` mega Hertz. What are the wave length of radiations from each station? |
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Answer» Station `I^(st) :lambda = (C )/(v)=(3xx10^(8)m sec^(-1))/(1120xx10^(3)sec^(-1)) = 267.86m` Station `II^(nd) : lambda = (C )/(v) = (3xx10^(8)msec^(-1))/(98.7xx10^(6)sec^(-1))=3.0395m` |
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| 1939. |
AIR service on Vividh Bharati is transmitted on `219m` band. What is its transmission frequency in Hertz ?A. `1.37xx10^(6)Hz`B. `1.37xx10^(5)Hz`C. `1.37xx10^(7)Hz`D. `2.74xx10^(5)Hz` |
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Answer» Correct Answer - 1 Given `lambda=219m` Thus, `v=(c )/(lambda)" "or " "v=(3xx10^(8))/(219)=1.37xx10^(6)Hz` |
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| 1940. |
The change in molar enrgy noticed during atomic transimission of 5 hertz frequency from a monoatomic molecule :A. `33xx . 1 xx 10^(-34) J`B. ` 1.99xx 10^(-9) J`C. `30.1xx 10^(-9)J`D. None of these |
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Answer» Correct Answer - B `E= N. h.v` `=6. 023 xx 10^(23) xx 6.624 xx 10^(-34) xx 5` ` = 1.99 xx 10^(-9) j//mol`. |
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| 1941. |
The angular momentum of an electron in a orbital is given as …………A. `nh/(2pi)`B. `h/(2pi) xx sqrt ( l (l+1))`C. `n h/( 2 pi`D. None of these |
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Answer» Correct Answer - B It is expression to represent angular momentum of an elecrtron is an orbital . |
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| 1942. |
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength ` 6800` Å . Calculate threshold frequency `( v_0)` and work function `(W_0)` of the metal. |
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Answer» `E// "photon" = (hc)/(lambda) = ( 6.626 xx 10^(-34) xx 3.0 xx 10^8)/( 6800 xx 10^(-28)` . `= 2. 92 xx 10^(19) J` `E//" photon" = 2.92 xx 10^(-19) J` = work function (The ejection has zero velocity ) Threshold frequency `= c/(lambda) = ( 3 xx 10^8)/( 6800 xx 10^(-10))` ` = 4. 41 xx 10^(14) sec ^(-1)`. |
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| 1943. |
What is the wavelength of light emitted when the electron of a hydrogen atom undergoes a transition from an energy level with `n= 4` to an energy level with `n = 2`? .What is the colour corresponding to this wavelength ? |
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Answer» The frequency (v) of radiation when an electron jumps from a higher orbit `v_(2)` to a lower orbit `v_(1)` is given by `v = 3.29 xx 10^(15)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` Therefore frequency v when an electron jumps from `n = 4 "to" n = 2` is given by `v = 3.29 xx 10^(15)((1)/((2)^(2)) -(1)/((4)^(2)))` `3.29 xx 10^(14) s^(-1)` we also know `v lambda = c` `or lambda = (c )/(v) = (3 xx 10^(6) ms^(-1))/(6.85 xx 10^(14) s^(-1))` `= 4.85 xx 10^(-7) m =- 485 nm` THe colour is blue |
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| 1944. |
What is the wavelength of light emitted when the electron of a hydrogen atom undergoes a transition from an energy level with `n= 4` to an energy level with `n = 2`? .What is the colour corresponding to this wavelength ? (Given `R_(H) = 109677 cm^(-2))` |
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Answer» According to gbalmer equation Wavelength `(bar v) = R_(H) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))] cm^(-1)` `n_(1) = 2,n_(2) = 4, R_(H) = 109677 cm^(-1)` substituting all the value , we get `bar v = 109677 [(1)/((2)^(2)) - (1)/((4)^(2))] cm^(-1) = (10977 xx 3)/(16) cm^(-1)` `or lambda = (1)/(bar v ) = (16)/(109677 xx 3) cm = 486 nm ` The wavelength corresponds to blue colour |
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| 1945. |
What is the wavelength of light emitted when the electron of a hydrogen atom undergoes a transition from an energy level with `n= 4` to an energy level with `n = 2` ? What is the colour corresponding to this wavelength? |
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Answer» ` Delta E =E_4 - E_2 = (-13.6)/(4^2) - ( - ( 13. 6)/2^2) ( therefore E_n= (E_1)/(n^2))` ` Delta E= ( + 13. 6 xx 3)/( 16) eV` ` therefore (12375)/(lambda) = (13. 6 xx 3)/(16)` ` ( lambda "in " Å )` `therefore lambda = 4852 Å = 485.2 nm`. |
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| 1946. |
The work function `(phi)` of some metals is listed below. The number of metals which will show photoelectric effect when light of `300` nm wavelength falls on the metal is`((Me t a l ,Li, Na , K , Mg, Cu, Ag , Fe , Pt, W),( phi(eV), 2.4, 2.3, 2.2, 3.7, 4.8, 4.3, 4.7, 6.3, 4.75))` |
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Answer» Correct Answer - `4.14eV` |
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| 1947. |
The numer of waves made by a bohr electron in an orbit of maximum magnetic quantum number +2 is:A. 3B. 4C. 2D. 1 |
| Answer» Correct Answer - A | |
| 1948. |
In the Rutherford experiment, `alpha`-particles are scattered from a nucleus as shown. Out of the four paths, which path is not possible? A. ElectronB. ProtonC. AtomD. Nucleus |
| Answer» Correct Answer - D | |
| 1949. |
What will be de-Broglie wavelength of an electron moving with a velocity of `1.2xx10^5 "ms"^(-1)`A. `6.071xx10^(-9)`B. `3.133xx10^(-37)`C. `6.626xx10^(-9)`D. `6.018xx10^(-7)` |
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Answer» Correct Answer - A `lambda=h/p , p=mv` `lambda=h/"mv"=(6.62xx10^(-34))/(9.1xx10^(-31)xx1.2xx10^5)` `lambda=6.071xx10^(-9)`m |
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| 1950. |
An electron having velocity `2xx10^(6)m//s` has uncertainty in kinetic energy is `(6.66)/pixx10^(-21) J`, then calculate the uncretainty in position (in Angstrom ,Å) of the electron .[Given `:h=6.60xx10^(-34) J-sec]` |
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Answer» `KE=(1)/(2) mv^(2)" "V=2xx10^(6)` `d(KE)=mvdv` `dv=(d(KE))/(mv)" "......(1)" "But" "Delta"x"=(h)/(4pimDeltav)" "....(2)` `Delta"x"=(h)/(4pim(d(KE))/(mv))" " , " "Delta"x"=(6.62xx10^(-34)xx2xx10^(6))/(4pixx6.62/(pi)xx10^(-21))=500Å` |
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