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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.
| 1. |
The ratio between the neutrons present in nitrogen atom and silicon atoms with number 14 and 28 is :A. `7:3`B. `3:7`C. `1:2`D. `2:1` |
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Answer» Correct Answer - B `n = A -Z` |
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| 2. |
Which one of the following sets of ions represents the collection of isoelectronic species?A. `Na^(+),Mg^(2+),Al^(3+),Cl^(-)`B. `Na^(+),Ca^(2+),Sc^(3+),F^(-)`C. `K^(+),Cl^(-),Mg^(2+),Sc^(3+)`D. `K^(+),Ca^(2+),Sc^(3+),Cl^(-)` |
| Answer» Correct Answer - D | |
| 3. |
If `I_(0)` is the threshold wavelength for photoelectric emission, 1 the wavelength of light falling on the surface of a metal and m is the mass of the electron, then the velocity of ejected electron is given byA. `[(2h)/(m)(lambda - lambda)]^(1//2)`B. `[(2hc)/(m)(lambda_(0)-lambda)]^(1//2)`C. `[(2hc)/(m){(lambda_(0)-lambda)/(lambda lambda_(0))}]^(1//2)`D. `[(2hc)/(m){(1)(lambda_(0))-(1)/(lambda)}]^(1//2)` |
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Answer» Correct Answer - C `(hc)/(lambda) = h(c)/(lambda_(0)) +(1)/(2)mv^(2), hc[(1)/(lambda)-(1)/(lambda_(0))] = (1)/(2)mv^(2)` `V = ((2hc)/(m)[(lambda_(0)-lambda)/(lambda_(0)lambda)])^(½)` |
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| 4. |
If the threshold wavelength `(lambda_(0))` for spection of electron from metal is `350nm `then work function for the photoelectric emission isA. `1.2 xx 10^(-18) J`B. `1.2 xx 10^(-20) J`C. `6 xx 10^(-29) J`D. `6 xx 10^(-12) J` |
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Answer» Correct Answer - B Work function = Threshold energy `= hv_(0) = (hc)/(lambda_(0))` ` = (6.6 xx 10^(-34) J s xx 3 xx 10^(8) m)/(330 xx 10^(-9) m) = 6.6 xx 10^(-29) J` |
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| 5. |
Which one of the following sets of ions represents a colloection of isoelectronic species ?A. ` K^+ , Cl^- , Ca^(2+) , Sc^(3+)`B. ` K^+ , Cl^- , Ca^(2+) , Sc^(2+), Sc^(3+)`C. `N^(3-), O^(2-) F^- , S^(2-)`D. `Li^+ , Na^+ , Mg^(2+) , ca^(2+)`. |
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Answer» Correct Answer - A Each has ` 18` electrons. |
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| 6. |
If `lambda_(0)` is the threshold wavelength for photoelectric emission. `lambda` wavelength of light falling on the surface on the surface of metal, and `m` mass of electron. Then de Broglie wavelength of emitted electron is :-A. `[(h(lambdalambda_(0)))/(2mc(lambda_(0)-lambda))]overset(1)(2)`B. `[(h(lambda_(0)-lambda))/(2mclambdalambda_(0))]overset(1)(2)`C. `[(h(lambda-lambda_(0)))/(2mclambdalambda_(0))]overset(1)(2)`D. `[(hlambdalambda_(0))/(2mc)]overset(1)(2)` |
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Answer» Correct Answer - A `m_(n)` = mass of neutron , `m_(p)` = mass of proton `(m_(n))/(2)" "2m_(p)` atmoic mass `implies (m_(n)+m_(p))" "[m_(n)~=m_(p)] implies (8+6)=14m_(p)` atomic mass `implies (4+12)=16m_(p)` `%` increase `=(16-14)/(14)xx100=14.28%` |
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| 7. |
If the threshold wavelength `(lambda_(0))` for ejection of electron from metal is `350 nm ` then work function for the photoelectric emission isA. `1.2 xx 10^-18 J`B. `1.2 xx 10^-20 J`C. `6 xx 10^-19 J`D. `6 xx 10^-12 J` |
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Answer» Correct Answer - C ( c) `W = h ( C)/(lamda_0)` :. `W = (6.6 xx 10^-34 xx 3 xx 10^8)/(330 xx 10^-9)` =`(6.6 xx 3 xx 10^-18)/(33)` =`0.6 xx 10^-18` =`6 xx 10^-19 J`. |
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| 8. |
Elements up to atomic number `103` have been synthesized and studied. If a newly discovered element is found to have an atomic number `106`, its electronic configuration will beA. `[Rn]5 f^14 , 6d^4, 7s^2`B. `[Rn]5 f^14 , 6d^1, 7s^2 7p^3`C. `[Rn]5 f^14 , 6d^6, 7s^0`D. `[Rn]5 f^14 , 6d^5, 7s^1` |
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Answer» Correct Answer - D `Unh_106 =[Rn]5f^14 , 6d^5, 7s^1` |
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| 9. |
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [`a_0` is Bohr radius] :A. `(h^(2))/(4 pi^(2) ma_(0)^(2))`B. `(h^(2))/(16 pi^(2) ma_(0)^(2))`C. `(h^(2))/(32 pi^(2) ma_(0)^(2))`D. `(h^(2))/(64 pi^(2) ma_(0)^(2))` |
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Answer» Correct Answer - C |
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| 10. |
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [`a_0` is Bohr radius] :A. `h^2/(4pi^2 ma_0^2)`B. `h^2/(16pi^2 ma_0^2)`C. `h^2/(32pi^2 ma_0^2)`D. `h^2/(64pi^2 ma_0^2)` |
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Answer» Correct Answer - C `mv(4a_0)=h/pi` so, `v=h/(4mpia_0)` so, `KE=1/2mv^2=1/2m.(h^2)/(16m^2pi^2a_0^2)=(h^2)/(32mpi^2 a_0^2)` |
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| 11. |
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [`a_0` is Bohr radius] :A. ` h^2/( 4 pi ^2 ma_0^2)`B. ` h^2/( 16pi ^2 ma_0^2)`C. ` h^2/( 32 pi ^2 ma_0^2)`D. ` h^2/(6 4 pi ^2 ma_0^2)` |
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Answer» Correct Answer - C For Bohr orbit . ` K.E, = 1/2 mu^2` and `mur_n = (nh)/(2pi) ro u = (nh)/( 2pi mr_n)` ` :. K.. 1/2 m = (n^2 h^2 )/( 4pi^2 m^2 r_n^2) = (h^2 h^2)/(8 pi^2 mr_n^2)` For (II) Bohr orbit , n=2 `:. r_n = a_0 xx n^2 =4a_0 (a_0 = "Bohr radius" )` Now ` K.E. = ((2)^2 h^2)/(( 8 ppi^2 m 9 4 a_0)^2) = h^2/(32 pi^2 ma_0^2)`. |
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| 12. |
Ther ratio of the radii of three Bohr orbit isA. `1 : 05 : 3`B. `1 : 2 : 3`C. `1 : 4: 9`D. `1 : 8 : 27` |
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Answer» Correct Answer - C ( c) `r prop n_2` `n = 1, 2, 3….` so ratio `= 1 : 4 : 9`. |
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| 13. |
If the radius of first Bohr orbit be `a_0`, then the radius of the third orbit would be-A. `3 xx a_0`B. `6 xx a_0`C. `9 xx a_0`D. `1//9 xx a_0` |
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Answer» Correct Answer - C ( c) `r = 0.529 (n^2)/(Z) Å` `n = 1 " " Z = 1` so `r = 0.529 Å = a_0` "Therefore for third orbit" `r = 0.529 xx 9` or `r = 9 xx 0`. |
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| 14. |
The magneit moment of a transitio metal is ` sqrt(15)` B.M . Fin dout the mubner of unpaired electrons in it . |
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Answer» Correct Answer - 3 `:.` Magetic moment `= sqrt(n( n + 2 )) B.M`. (wheree (n) is unpaired electron ) ` sqrt(15) = sqrt(n(n + 20)) rArr n = 3` . |
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| 15. |
The magneit moment fo electron in an atom ( excluding orbitial magnetci moment ) is given by :A. ` sqrt ( n ( n+ 2 ))` Bohr MagnetonB. ` sqrt ( n ( n+ 1 ))BM`C. ` sqrt ( n ( n+ 3 ))BM`D. None fo the above |
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Answer» Correct Answer - A The formula for magnetic moment of an atom . |
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| 16. |
Which set is correct for an electron in `4f -orbitial` ?A. `n = 3 l = 1` m_1 = -2 m_s = + .^1//_2`B. `n = 4 l = 4 m_l = -4 m_s = - .^1//_2`C. `n = 4 l = 3 m_l = + 1 m_s = + .^1//_2`D. `n = 4 l = 3 m_l = + 4 m_s = + .^1//_2` |
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Answer» Correct Answer - C ( c) For `4f` orbital `n = 4, 1 = 3` For `1 = 3, m ne 4`. |
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| 17. |
What are the numbers of nodes present in : (a) 1 s, (b) 2s, (c ) 2 p, (d) 3 p, orbitial s? |
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Answer» (a) Zero. (b) One spherical node . ( c) One angular node , (d ) Two angular node |
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| 18. |
Principal azimuthal , and magnetic quantum numbers are respetively related toA. size, shape and orientationB. shape, size and orientationC. size, orientation and shapeD. None of the above |
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Answer» Correct Answer - A (a) `n, 1` and `m` are related to size, shape and orientation respectively. |
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| 19. |
Principal azimuthal , and magnetic quantum numbers are respetively related toA. Size,orbital,and shapeB. size, shape, and orientationC. shape, size and oricutationD. None of these |
| Answer» Principal quantum number gives size azimaller gives shape red magnetic orientation | |
| 20. |
`{:(,"ColumnI",,"ColumnII"),((A),"The d-orbital which has two angular nodes",(P),3d_(x^(2)-y^(2))),((B),"The d-orbitial with two nodal surfaced from conce",(Q),3d_(s^2)),((C),"The orbital without angular node",(R),4f),((D),"The orbital which has three angular nodes",(S),3s):}` |
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Answer» Correct Answer - A-p, B-q, C-s, D-r no of angular nodes = l |
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| 21. |
The permissible solution to the scheodinger wave an idea of ………. Quantum numberA. 4B. 2C. 3D. 1 |
| Answer» Correct Answer - `3(n,l,m)` | |
| 22. |
The ratio of the differrence between the first and second Bohr orbit energies to that between second and third Bohr orbit energies isA. `1//3`B. `27//5`C. `9//4`D. 4//9` |
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Answer» Correct Answer - B `1 E_n = E_1 /n^2` , Find `((E_2 - E_1))/(( E_3 -E_2))`. |
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| 23. |
If the radius of the second Bohr of hydrogen atom is `r_(2)` the radius of the third Bohr orbit will beA. `(4)/(9) r_2`B. `4 r_2`C. `(9)/(4) r_2`D. `9 r_2` |
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Answer» Correct Answer - C ( c) `r = (n^2 h^2)/(4 pi^2 m Ze^2)` `therefore (r_2)/(r_3) = (2^2)/(3^2)` `therefore r_(3) = (9)/(4) r_2`. |
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| 24. |
If the radius of the second Bohr of hydrogen atom is `r_(2)` the radius of the third Bohr orbit will beA. `(4)/(9)r_(2)`B. `4r_(2)`C. `(9)/(4)r_(2)`D. `9r_(2)` |
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Answer» `r = (n^(2)h^(2))/(4pi^(2)mZe^(2))` `:. (r_(2))/(r_(3)) = (2^(2))/(3^(2))` `:. V_(3) = (9)/(4)r_(2)` |
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| 25. |
The velocity of an electron in the second Bohr orbit of an element is `1.1 xx 10^(6) s^(-1)` Its velocity in the third orbit isA. `3.3 xx 10^(6) m s^(-1)`B. `2.2 xx 10^(6) m s^(-1)`C. `7.333 xx 10^(5) m s^(-1)`D. `3.66 xx 10^(5) m s^(-1)` |
| Answer» `7.333 xx 10^(5) m s^(-1)` | |
| 26. |
An energy of `68eV` is required to excite a hydrogen like atom from its second Bohr orbit to the third. The nuclear charge is Ze. Find the value of Z, the kinetic energy of the electron in the first Bohr orbit and the wavelength of the radiation required to eject the electrons from the first Bohr orbit to infinity. |
| Answer» Correct Answer - `6;489.6eV,25.28A^(@)` | |
| 27. |
The ratio of the differrence between the first and second Bohr orbit energies to that between second and third Bohr orbit energies isA. `(1)/(2)`B. `(1)/(3)`C. `(27)/(3)`D. `(5)/(27)` |
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Answer» `Delta E = - 2.18 xx 10^(-18)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) J" atom" ^(-1)` `Delta E_(2 rarr 1) = - 2.18 xx 10^(-18)((1)/(1^(2)) - (1)/(2^(2)))` ` = - 2.18 xx 10^(-18)((3)/(4))` `Delta E _(3 rarr 2)` = - 2.18 xx 10^(-18) ((5)/(36))` `:. (Delta E_(2-1))/(Delta E_(1-2)) = (3)/(4) xx (36)/(5) = (27)/(5) :. (c )` |
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| 28. |
The electrons, identified by quantum number n and l i. `n = 4,l=1` ii. `n = 4, l= 0` iii. `n = 3 , l= 2 ` iv. `n= 3 , l = 1` Can be palced in the order of increasing energy from the lowest to highest,itsA. `(iv) lt (ii) lt (iii) lt (i)`B. `(ii) lt (iv) lt (i) lt (iii)`C. `(i) lt (iii) lt (ii) lt (iv)`D. `(iii) lt (i) lt (iv) lt (ii)` |
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Answer» Correct Answer - A (a) The two guiding rules to arrange the various orbitals in the increasing energy are : (i) Energy of an orbital increases with increase in the value of `n + 1` (ii) Of orbitals having the same value of `n +1`, the orbital with lower value of `n` has lower energy. Thus for the given orbitals, we have (i) `n + 1 = 4 + 1 = 5` (ii) `n + 1 = 4 + 0 = 4` (iii) `n + 1 = 3 + 2 = 5` (iv) `n + 1 = 3 + 1 = 4` Hence, the order of increasing energy is. `(iv) lt (ii) lt (iii) lt (i)`. |
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| 29. |
Elements up to atomic number `103` have been synthesized and studied. If a newly discovered element is found to have an atomic number `106`, its electronic configuration will beA. `[Rn] 5 f^14, 6 d^4, 7 s^2`B. `[Rn] 5 f^14, 6 d^1, 7 s^2 7 p^3`C. `[Rn]5 f^14, 6 d^6, 7 s^0`D. `[Rn] 5 f^14, 6d^5, 7 s^1` |
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Answer» Correct Answer - D (d) `UnH_(106) = [Rn] 5f^14, 6d^5, 7 s^1`. |
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| 30. |
The only element in the hydrogen atom resides under ordinary condition on the first orbit .When energy is supplied the element move to hjgher energy ornbit depending on the lower of energy absioerbed .When this electron to may of the electron return to any of the lower orbits, it emit energy Lyman series is formed when the electron to the lowest orbit white Balmer series ids formed when the electron returns to the second orbit similar Paschen Brackett, and Pfund series are formed when electron return to the third fourth , and fifth arbit from highest energy orbits, respectively Maximum number of liner produced is equal when as electron jumps from nth level to ground level is equal to `(n(n - 1))/(2)`If teh electron comes back from the energy level having energy `E_(2)` to the energy level having energy `E_(1)` then the difference may be expresent in terms of energy of photon as `E_(2) - E_(1) = Delta E, lambda = hc//Delta E` Since h and c are constants `Delta E` coresponding to definite energy , thus , each transition from one energy level to unother will produce a light of definite wavelem=ngth .This isd actually observed as a line in the spectrum of hydrogen atom Wave number of line is given by the formula `bar v = RZ^(2)((1)/(n_(1)^(2))- (1)/(n_(12)^(2)))`Where R is a Rydherg constant The wave number of electromagnetic radiation emitted during the transition of elecvtron in between the two levels of `Li^(2+)` ion whose pricipal quantum numbner sum is `4` and difference is `2` isA. `3.5R_(H)`B. `4R_(H)`C. `8R_(H)`D. `(8)/(9)R_(H)` |
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Answer» Correct Answer - C `8R_(H)` |
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| 31. |
Which of the following is false?A. Breaker spectral series for which `n_(1) = 4` and `n_(2) = 5,6,7….` lies in the infrared regaion of the electromagnetic radiationB. The orbitals `3d_(x^(2)` is symmetrical sbout z-axisC. The orbital `3d_(xy)` has no probability of finding electron along x-and y-axisD. The orbital `3d_(x^(2) - y^(2)` has probabilityy of linding electron along x- and y-axis |
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Answer» Correct Answer - D Bosons does not follow paupli exclusion principle. |
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| 32. |
Which of the following is false?A. Pfund spectral series for which `n_(1) = 5` and `n_(2) = 6,7…..` lies the infrared region of the electronetic radationB. Visible region of electromagnetic radiation has wavelength from `400 nm to 800 nm`C. Balmer spectral series lies in the visible proton of the electromagnetic radiationD. Lyman series lies in the visible protion of the electronetic radiation |
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Answer» Correct Answer - D d. Lyman spectral series lies in the ultraviolet region |
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| 33. |
Which of the following is true ?A. The half -filled and filled elkectronic configuration are less stable than the other congfiguration having the same number of electronB. The symbols s for the orbitals having `l = 0` has iots origim=n from the term spherical symmetricalC. The insreasing order for the value of e//m (charge //mass) for eletron (e ) proton (p) neatron (n) and alpha partickle (u) is `n it alpha it p it e`D. The energy of photon having wavelength `800 nm` is larger than having `400 nm` |
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Answer» Correct Answer - C a. The half -filled and fully -filled electronic configuration are stable it is due to the large exxharge energy b . The symbol s stand for shape-a term used in the characterassion of spectral lines , The symbols p,d,and f stand for principle , differeus , and fundamental respectively d. The expression is `E = hc//lambda`Hence energy and wavelength are inversaly reated |
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| 34. |
The shape of orbitals are related to the ratio of principal quantum number (n) to substiary quantum number (k,a modifacation of Bohr-sommerfield theory ).The value of k for any shell has a value ranging betwe3en n to l .The amximum value for k is given for x sub-shell white k becomes with p, d,f........ repectively upto minimum value If n is the major axis and k is th e minor axis , then `n//k = 1` for circular shape white `n//k gt 1` for elliptical shape The ratio of `n//k = 2` does not related toA. 2pB. 4dC. 6fD. 2s |
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Answer» Correct Answer - D 25 |
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| 35. |
The shape of orbitals are related to the ratio of principal quantum number (n) to substiary quantum number (k,a modifacation of Bohr-sommerfield theory ).The value of k for any shell has a value ranging betwe3en n to l .The amximum value for k is given for x sub-shell white k becomes with p, d,f........ repectively upto minimum value If n is the major axis and k is th e minor axis , then `n//k = 1` for circular shape white `n//k gt 1` for elliptical shape Which value of n and k suggest about the shape of `3s` orbitsl?A. 3,2B. 1,1C. 3,0D. 3,3 |
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Answer» Correct Answer - D 3,3 |
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| 36. |
The shape of orbitals are related to the ratio of principal quantum number (n) to substiary quantum number (k,a modifacation of Bohr-sommerfield theory ).The value of k for any shell has a value ranging betwe3en n to l .The amximum value for k is given for x sub-shell white k becomes with p, d,f........ repectively upto minimum value If n is the major axis and k is th e minor axis , then `n//k = 1` for circular shape white `n//k gt 1` for elliptical shape Which is correct according to the increasing elliptical number of sub-shell ?A. `2s lt5p lt 3plt 4d`B. `4d lt 2s lt 5plt 3p`C. `4d lt2s lt 3plt 5p`D. `3p lt4d lt 2slt 5d` |
| Answer» Correct Answer - A | |
| 37. |
What is likely to be principal quantum number for a circular orbit of diameter `20nm` of the hydrogen atom if we assume Bohr orbit be the same as that represented by the principal quantum number?(a). `10`(b). `14`(c). `12`(d). `16`A. 10B. 14C. 12D. 16 |
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Answer» Correct Answer - B |
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| 38. |
What is likely to be principal quantum number for a circular orbit of diameter `20.6nm` of the hydrogen atom. If we assume Bohr orbit to be the same as that represented by the principal quantum number?A. 10B. 14C. 12D. 16 |
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Answer» Correct Answer - B `r_(n) = 0.529 xx n^(2)A^(@)`, diameter =` 2r, r =("diameter")/(2)` |
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| 39. |
What is likely to be principal quantum number for a circular orbit of diameter `20nm` of the hydrogen atom if we assume Bohr orbit be the same as that represented by the principal quantum number?A. `10`B. `14`C. `12`D. `16` |
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Answer» Correct Answer - B `d=20nm` `r=(20)/(2)=10nm=100Å` `therefore r=0.529xx(n^(2))/(Z)A^(@)` For `H` atom `Z=1` `100=0.529xxn^(2) " "n=14` |
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| 40. |
Which ofthe following curves may represent the energy of electron in hydrogen atom as a function of principal quantum number n:A. B. C. D. |
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Answer» Correct Answer - A `V prop (1)/(n)` |
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| 41. |
Which of the following curves may represent the radius of orbit `(r_(n))` in H-atoms as a function of principal quantum number (n)A. B. C. D. None of these |
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Answer» Correct Answer - B `r prop (n^(2))/(z)` |
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| 42. |
The quantum number `+ 1//2` and `-1//2` for the electron spin representA. rotation of the electron in clockwise and anticlockwise directions respectivelyB. rotation of the electron in anticlockwise and clockwise direction respectively.C. magnetic momentum of electron pointing up and down respectivelyD. two quantum mechanical spin states which have no classical anologues. |
| Answer» Correct Answer - D | |
| 43. |
A compound of vanadium has a magnetic moment of `1.73BM` Work out the electronic configuration of vanadius in the compound |
| Answer» Correct Answer - A::B::C::D | |
| 44. |
A compound of vanadium has a magnetic moment of `1.73BM` . Work out the electronic configuration of vanadium ion in the compound. |
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Answer» No. of unpaird electron are given by Magnetic moment `= sqrt ( [n ( n + 20 )]` ( where (n) is no , of unpaired electrons ) or ` 1. 7 3 = sqrt ( [n (n + 2)])` or ` 1. 73 xx 1. 73 = n^2 + 2 n :. N=1` Now vanadium atom must have one unpaired electron and thus its configuration is ` ._(23) V^(4+) : 1s^2 , 2s^2 , 2s^2 . 2p^6 . 3 s^2 3p^6 3d^1`. |
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| 45. |
Statement : wavelength of (I) line of Humphry series is more than (I) line of Lyman series in H-atom Explanation : `Delta E = (hc)/( lambda)`.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - A (a) `Delta E_1 = E_6 - E_5` in Humphrey series and `Delta E = E_2 - E_1` for Lyman series `because Delta E gt Delta E_1` `lamda_(Lyman) lt lamda_(Humphry)`. |
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| 46. |
Which of the following may represent the possbile quantum numbers for last electron of Ga.A. `3,1,+1,+(1)/2`B. `4,0,1+1,+(1)/(2)`C. `4,1,0,_(1)/(2)`D. `4,1,+1,+(1)/(2)` |
| Answer» Correct Answer - D | |
| 47. |
A compound of vanadium has a magnetic moment of `1.73BM`. Work out the electronic configuration of vanadium in the compoundA. `[Ar]3d^(1)`B. `[Ar]3d^(2)`C. `[Ar]3d^(3)`D. `[Ar]3d^(0)` |
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Answer» Correct Answer - A `sqrt(n(n+2))=1.73` `n(n+2)=3` `n=1` So vandium has 1 unparied electron and its configuration is`[A]3d^(1)`. |
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| 48. |
`A` light source of power 16 watts emits light of wavelength `310nm`. If all emitted photons are made to strike a metal plate of work function `1.5eV` then find out the magnetic of photo-current if `50%` of incident photons eject photoelectrons. |
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Answer» No. of photons per second from source `=("power")/("energy of photon")=(16J//sec)/((1240)/(310)xx1.6xx10^(-19)J) = 2.5xx10^(19) "photons/sec"` No. of photoelectron emitted `=(1)/(2)(2.5xx10^(19))` per sec. `("charge")/("time")` = photoelectron current `=(1)/(2)(2.5xx10^(19))xx1.6xx10^(-19)C//sec = 2` amp |
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| 49. |
The wavelength of the electron emitted by a metal sheet of work function 5 eV when photons from EMR of wavelength 62 nm strike the metal plate .A. `82.667 "Å"`B. 3.16 nmC. 0.316 nmD. `826. 67 "Å"` |
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Answer» Correct Answer - C |
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| 50. |
The eyes of certain member of the reptile family pass a single visual signal to the brain when the visual receptors are struck by photons of wavelength `850nm`. If a total energy of `3.15xx10^(14)J` is required to trip the signal. What is the minimum number of photons that must strike the receptor? |
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Answer» Correct Answer - `1.35xx10^(5)` `E=(hL)/(lambda) " "n=(3.15xx10^(-14)xx850xx10^(-9))/(6.62xx10^(-34)xx3xx10^(8))` `n=134.8xx10^(3)" "n=1.35xx10^(5)` |
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