InterviewSolution
Saved Bookmarks
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.
| 2001. |
Calculate de-Broglie wavelength when `e^(-)` is accelerated by the following voltage. (i) `750V` (ii) `300` volt |
| Answer» (i) `" "lambda=sqrt((150)/(V))Å" "=sqrt((150)/(750))Å=(1)/sqrt(5)Å " "(ii)" "lambda=sqrt((150)/(300))Å=(1)/(sqrt(2))Å` | |
| 2002. |
de Broglie wavelength of an electron after being accelerated by a potential difference of V volt from rest is :A. `lambda = (12.3)/(sqrth) "Å"`B. `lambda = (12.3)/(sqrtV) "Å"`C. `lambda = (12.3)/(sqrtE) "Å"`D. `lambda = (12.3)/(sqrtm) "Å"` |
|
Answer» Correct Answer - B |
|
| 2003. |
Assertion (A) :Angular momentum of `1s,2s,3s,` ets all have spectrical shape Reason (R ) : `1s,2s,3s,` ets all have spectrical shapeA. If both (A) and (R ) correct and (R ) is the correct explanation for (A)B. If both (A) and (R ) correct and (R ) is the correct explanation for (A)C. If (A) is correct but (R ) is incorrectD. If (A) is incorrect but (R ) is correct |
|
Answer» Correct Answer - A Reason is not the correct explanation |
|
| 2004. |
Calculate the shortest wavelength in H spectrum of Lyman series, when `R_(H) = 109677 cm^(-1)`. |
|
Answer» For lyman series `n_(1) = 1` For shortest `lambda `of Lyman series difference in two levels showing transition should be maximum (i.e.`n_(2) = prop` `:. (1)/(lambda) = R_(H) [(1)/(1^(2)) - (1)/(2^(2))] = 109678` `lambda = 911.7 xx 10^(-8) cm = 911.7 Å` For shortest `lambda `of Lyman series difference in two levels showing transition should be maximum (i.e.`n_(2) = 2` `:. (1)/(lambda) = R_(H) [(1)/(1^(2)) - (1)/(2^(2))] = 109678 xx (3)/(4)` `lambda = 1215.67 xx 10^(-8) cm = 1215.67 Å` |
|
| 2005. |
The de Broglie wavelength of an electron accelerated by an electric field of V volt is given by:A. `lamda=(1.23)/(sqrt(m))`B. `lamda=(1.23m)/(sqrt(h))`C. `(1.23)/(sqrt(V))nm`D. `lamda=(1.23)/(V)`. |
| Answer» Correct Answer - B | |
| 2006. |
Calculate the frequency of the spectrical line emitted when an electron in `n = 3` in H de-excited to the ground state `R_(H) = 109.737 cm^(-1)` |
|
Answer» `(1)/(lambda) = R_(H) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` `:.(1)/(lambda) = v=R_(H) .c[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` `= 109737 xx 3.0 xx 10^(19) [(1)/(1^(2)) - (1)/(3^(2))]` `= 2.92 xx 10^(15) s^(-1)` |
|
| 2007. |
If the uncertainnty in the position of a moving electron is equal to its de Broglie wavelength then moving its velocity will be completely anertain Explain . |
|
Answer» Given that the uncertainty in position `(delta x)` is equal to de Broglie wavelength `(lambda)i.e. Delta x = lambda ` But according to de Broglie equation `lambda = (h)/(mv) = (h)/(p)` `or Delta x = (h)/(p)` (Given condition ) According to Heisenbegy uncertainty principle `(Delta x) (Delta p) ge (h)/(4pi)` `((h)/(p)) Delta p ge (h)/(4pi)` `(Delta cancel p)/(cancel p) ge (cancelh)/(4pi xx cancel h)` `(Delta cancel p)/(cancel p) ge (1)/(4pi)` (constant) |
|
| 2008. |
The uncertainty in the momentum of a particle is `6.0 xx 10^(-2) kg m s^(-1)` .Calculate the uncertainty in the position |
|
Answer» Uncertainty in momentum `Delta p = 6 xx 10^(-2) kg m s^(-1)` According to Heisenbery uncertainty principle ` Delta x xx Delta p = (h)/( 4pi)` `Delta x = (h)/(Delta p xx 4pi)=(6.6 xx 10^(-34) kg m^(2)s^(-2))/(6 xx 10^(-2) kg m s^(-1) xx 4 xx3.14)` `= 8.76 xx 10^(-34) m ` The uncertainty in its position is `8.76 xx 10^(-34) m` |
|
| 2009. |
Increasing order of magnetic monent among the following species is `Na^(+),Fe^(+3),CO^(2+),Cr^(+2)`A. `Na^(+) lt CO^(2+) lt Cr^(+2) lt Fe^(3+)`B. `Fe^(3+) lt Cr^(2+)ltCo^(+2)ltFe^(3+) lt Na^(+)`C. `Na^(+)ltCr^(2+)ltCo^(+2)ltFe^(3+)`D. `Na^(+)ltFe^(3+)ltCr^(2+)ltCo^(+2)` |
|
Answer» Correct Answer - A `Na^(+)ltCo^(2+)ltCr^(2+)ltFe^(3+)` Higher the number of unpaired electron will be magnetic moment . |
|
| 2010. |
Wavelength of the Balmer H, line (first line) is `6565 Å`. Calculate the wavelength of `H_(beta)` (second line). |
|
Answer» Correct Answer - `4863 A^(@)` `n_(2)=3" "n_(1)=2` [first line] `n_(2)=4" "n_(1)=2` [second line] `(1)/(lambda)=R_(H)[(1)/(4)-(1)/(9)]` `(1)/(6565)Å=R_(H)[(1)/(4)-(1)/(9)]....(i)` `(1)/(lambda)=R_(H)[(1)/(4)-(1)/(16)]...(ii)" "((i))/((ii))` `(lambda)/(6565)=((5)/(36))/((3)/(16))=(5xx16)/(36xx3)" "lambda=4863Å` |
|
| 2011. |
The `lambda` of `H_(alpha)` line of the Balmer series is `6500 Å` What is the `lambda` of `H_(beta)` line of the Balmer series |
|
Answer» For `H_(alpha)`line of the balmer series `n_(1) = 2,n_(2) = 3` For `H_(beta)`line of the balmer series `n_(1) = 2,n_(2) = 4` `1/(lambda_(H_(alpha))) = R_(H) [(1)/(2^(2)) - (1)/(3^(2))]`……..(i) ` and `1/(lambda_(H_(beta))) = R_(H) [(1)/(2^(2)) - (1)/(4^(2))]`……..(ii) By equation (i) and (ii) we get `:. (lambda_(beta))/(lambda_(alpha))= ((1)/(4) - (1)/(9))/((1)/(4) - (1)/(16))` `:. lambda_(beta) = lambda_(alpha) xx [(80)/(108)] = 6500 xx (80)/(108) = 4814.8 Å` |
|
| 2012. |
Wavelenth of the Balmer `H_(alpha)` line is 6565 "Å". Calculate the wavelength (in "Å") of `H_(beta)` line of same hydrogen like atom. |
|
Answer» Correct Answer - 4863 |
|
| 2013. |
In the following six electronic configuration (remaining inner orbital are completely flled). Mark the correct optio(s). A. Order of spin multiplicity : `C-IIgtC-I=C-IIIgtC-IV`B. Stabiltiy order : `C-IIgtC-I & C-IVgtC-III`C. If `C-V` represents element A then `A^(2+)` acts as diamagnetic.D. `C-VI` violates all the three ruls of electron configuration |
| Answer» Correct Answer - A,B,D | |
| 2014. |
What is the energy (in eV) required to excite the electron from n=1 to n=2 state in hydrogen atom ? (n=principal quantum number )A. 13.6B. 3.4C. `17.0`D. 10.2 |
|
Answer» Correct Answer - D According to Bohr equation `E_2-E_1=13.6(1/n_2^2-1/n_1^2)` or , `DeltaE=13.6(1/n_1^2-1/n_2^2) =13.6 (1-1/4) =10.2 eV` |
|
| 2015. |
The wavelength of `H_(alpha)` line of Balmer series is `X Å` what is the `X of H_(beta)`line of Balmer seriesA. `X (108)/(80) Å`B. `X (80)/(108) Å`C. `(1)/(X) (80)/(108) Å`D. `(1)/(X) (108)/(80) Å` |
|
Answer» Correct Answer - B `H_(alpha)` line of Balmer series means first line of Balmer series `n_(1) = 2,n_(2) = 3` `bar v = (1)/(lambda_(alpha)) = R ((1)/(2^(2)) - (1)/(3^(2))) = (5R)/(36)` `:. Lambda_(alpha)= (36)/(5R) = X` `H_(beta)` line of Balmer series means second line of Balmer series `n_(1) = 2,n_(2) = 4` `bar v = (1)/(lambda_(beta)) = R ((1)/(2^(2)) - (1)/(4^(2))) = (3R)/(16)` `:. Lambda_(beta)= (16)/(3R) = X` Then `(16)/(3R) = (X xx 5R xx 16)/(36 xx 3R) = (80)/(108) Å` |
|
| 2016. |
Let the I.E. of hydrogen like species be 320 eV . Find out the value of quantum number having the energy equal to `-20` eV .A. n = 2B. n = 3C. n= 4D. n = 5 |
|
Answer» Correct Answer - C |
|
| 2017. |
Isotopes exhibits similarA. Physical propertiesB. Chemical propertiesC. Physical and chemicalD. Neither physical nor chemical properties |
|
Answer» Correct Answer - B isotopes have same electronic configuration |
|
| 2018. |
A light wavelength `12818 Å` is emitted when the electron of a hydrogen atom drop from fifth to third quantum level .Find the wqavelemngth of the photon emitted when electron falls from third to ground level |
|
Answer» Correct Answer - A::B::D `For n_(1) rarr n_(3)` `(1)/(lambda) = R[(1)/(n_(2)^(2)) - (1)/(n_(1)^(2))]` `(1)/(12818) = R[(1)/(9) - (1)/(25)]` `R = (9 xx 25)/(16 xx 12818)Å` `For n_(3) rarr n_(1)` `(1)/(lambda) = R[1/1^(2) - (1)/(3^(2))] =(9 xx 25)/(16 xx 12818)xx (8)/(9)` `:. lambda = 1025 Å` |
|
| 2019. |
Which of the following statement (s) are incorrect?A. Photons having energy 400 kJ will break 4 mole bonds of a molecule `A_(2)` where A-A bond dissociation energy is `100kJ//"mol".`B. Two bulbs are emitting light having wavelength `2000"Å" "and" 3000"Å"` respectively. If the bulbs A and B are 40 watt and 30 watt respectively then the ratio of no. of photons emitted by A and B per day is `1/2`.C. When an electron make transition from lower to higher orbit photon is emitted.D. 4 eV is sufficient to excite an `e^(-)` from ground state of H-atom. |
|
Answer» Correct Answer - a,b,c,d |
|
| 2020. |
In which transition, one quantum of energy is emitted -A. `n=4 rarr n=2`B. `n=3 rarr n=1`C. `n=4 rarr n=1`D. `n=2 rarr n=1` |
| Answer» Correct Answer - A::B::C::D | |
| 2021. |
Angular momentum for P-shell electron is :A. `(3h)/(pi)`B. zeroC. `(sqrt(2)h)/(2pi)`D. none of these |
|
Answer» Correct Answer - c |
|
| 2022. |
In which transition, one quantum of energy is emitted -(a). `n=4 rarr n=2`(b). `n=3 rarr n=1`(c). `n=4 rarr n=1`(d). `n=2 rarr n=1`A. `n=4 to n=2`B. `n=3 to n=1`C. `n=4 to n=1`D. `n=2 to n=1` |
|
Answer» Correct Answer - a,b,c,d |
|
| 2023. |
If in the hydrogen atom P.E. At `oo` is chosen to be 13.6 eV then the ratio of T.E. to K.E.. For `1^(st)` orbit of H-atom is |
|
Answer» Correct Answer - A `T.E=-13.6-13.6=0` `K.E.=13.6+13.6=27.2` `(TE)/(KE)=0` |
|
| 2024. |
The energy of an electron in the first Bohr orbit is `-13 eV` . The energy of `Be^(3+)` in the first excited state is :A. `-30.6 eV`B. `-40 .8 eV`C. `-54.4` eVD. `+40.8` eV |
|
Answer» Correct Answer - C |
|
| 2025. |
How do the energy gaps between successive electron energy levels in an atom very from low to high n values ?A. All energy gaps are the same .B. The energy gap decreases as n increases.C. The energy gap increases as n increases.D. The energy gap changes unpredictably as n increases . |
|
Answer» Correct Answer - B |
|
| 2026. |
The magnitude of the spin angular momentum corresponding to an electron in Balmer transition inside a hydrogen atom cam be:A. `S=sqrt(s(s+1))(h)/(2pi)`B. `S=s(h)/(2pi)`C. `S=sqrt((3)/(2))xx(h)/(2pi)`D. `S=+-(1)/(2)xx(h)/(2pi)` |
|
Answer» Correct Answer - A::C Spin angular momentum`=sqrt(s(s+1))(h)/(2pi)` `S=sqrt((1)/(2)((1)/(2)+1))(h)/(2pi)=(sqrt(3))/(2)xx(h)/(2pi)` |
|
| 2027. |
The magnitude of the spin angular momentum corresponding to an electron in Balmer transition inside a hydrogen atom cam be:A. `S= sqrt(s(s+1))(h)/(2pi)`B. `S=s(h)/(2pi)`C. `S=(sqrt3)/(2) xx (h)/(2pi)`D. `S=+-1/2xxh/(2pi)` |
|
Answer» Correct Answer - a,c |
|
| 2028. |
The decreasing order of energy of the `3d,4s,3p,3s` orbital is :-A. `3dgt3sgt4sgt3p`B. `3sgt4sgt3pgt3d`C. `3dgt4sgt3pgt3s`D. `4sgt3dgt3sgt3p` |
|
Answer» Correct Answer - C The value of `l=0` to `(n-1)` Number of electron for given value of `l=2 (2l+1)` hence `underset(l=0)overset(l=(n-1))(sum)2(2l+1)` |
|
| 2029. |
The energy corresponding to second line of Balmer series for hydrogen atom will be :-A. 12.1 eVB. 1.89 eVC. 2.55 eVD. 13.6 eV |
| Answer» Correct Answer - 3 | |
| 2030. |
The radial wave function for 1 s electron in H-atom is `R=(2)/a_(0)^(3//2)e^(-r//a_(0))` where `a_(0)`=radius of `1^(st)` orbit of H-atom . The ratio of probablitiy of `1^(st)` electron in hyrogen atom at distance `a_(0)` from nucleus to that at distance `a_(0)/2` from nucleus.A. equalB. `(1)/etime`C. `(4)/etime`D. `(e)/4time` |
|
Answer» Correct Answer - C probability of finding and electron at a particular distance=`4pi_(2) R^(2)` `P_(1)=4pir^(2)R^(2)=4pir^(2)xx(4)/(a_(0)^(3))e^(-2r//a_(0))` `at" "a_(0)" "p_(1)=4pia_(0)^(2)xx(4)/(a_(0)^(3))e^(-2a_(0)//a_(0))` `at" "a_(0//2)" "p_(2)=4pi(a_(0)^(2))/(4)xx(4)/(a_(0)^(3))e^((-2a_(0))/(2a_(0)))` `at" "a_(0//2)" "p_(2)/(p_(2))=(4pia_(0)^(2)xx(4)/(a_(0)^(3))e^((-2a_(0))/(2a_(0))))/(4pia_(0)^(2)/(4)xx(4)/(a_(0)^(3))e^((-2a_(0))/(2a_(0))))` `(p_(1))/(p_(2))=(e^(-))/((1)/(4)e^(-1))=(4)/(e)` |
|
| 2031. |
If the velocity of an electron in `I^(st)` orbit of H is V, then what will be the velocity in `4^(th)` orbit of `Be^(+3)` :-A. VB. `(V)/(4)`C. 16 VD. 4 V |
| Answer» Correct Answer - 1 | |
| 2032. |
Assertion (A) : The first IE of Be is greater than that of B Reason (R ) : `2p` orbitals is lower in energy than `2s`A. If both (A) and (R) are correct and (R) is the correct reason for (A).B. If both (A) and (R) are correct but (R) is not the correct explanation for (A).C. If (A) is correct but (R) is incorrect.D. If (A) is incorrect but (R) is correct. |
| Answer» Correct Answer - C | |
| 2033. |
Magnetic moments of `V(Z = 23), Cr(Z = 24), Mn (Z = 25)` are x,y,z. HencE:A. `x=y=z`B. `xltyltz`C. `xltzlty`D. `zltyltx` |
| Answer» Correct Answer - 3 | |
| 2034. |
Which of the following ie//are posssible ?A. 3fB. 4dC. 2dD. 3p |
|
Answer» Correct Answer - B::D Angular quantum number (l) may have value less then the principal quantum number i.e. a. `4f n = 3,l= 3` b . `2d n = 2, d= 2` |
|
| 2035. |
Which of the following statement (s) are wrong.A. one mol of photons having ebnergy 400 KJ/mol will always breaks 4 male bonds of a molecule `A^(2)` where A-A bond dissocitaion energy id is 100 KJ/mol.B. Two bubls are emitting light having wavelenght ` 2000Å` ans `3000Å` respectively . If the bubls A and B ar 40 watt and 300 watt respectively then the ratio of number of photons emitted by A and B per dsy is `1//2`.C. When an electron make transitional from higher to lower orbit, photons(s) is /are emitted.D. Ratio of time period of electron in H-atom is `1^(st)` and `3^(rd)` level is `(9)/(1). |
|
Answer» Correct Answer - A::B::D (A) One mole photons will break one bond (b) `(n_(A))/(n_(B))=(40xx2000)/(30xx3000)=(8)/(9)` (D) Time period `n^(3)` `(T_(1))/(T_(3))=(1)/(9)` |
|
| 2036. |
Magnetic moments of `V(Z = 23), Cr(Z = 24), Mn (Z = 25)` are x,y,z. HencE:A. `x = y = z`B. `x lt y lt z`C. `x lt z lt y`D. `z lt y lt x` |
|
Answer» Correct Answer - C `mu = sqrt(n(n+2))` |
|
| 2037. |
The quantum yield for decomposition of HI id 0.2 .In an experiment 0.01 moles of HI are decomposed . Find the number of photons absorbed. |
|
Answer» Correct Answer - `3xx10^(22)` `(0.2n)/(Na)=0.01`mole" "`(0.2xxn)/(1+128)=0.01` `(0.2xxn)/(10xx127)=(1)/(100)" "2xxn=(127)/(10)` `n=(127)/(10xx2)=(12.7)/(2)=6` No. of protons `=(6xx10^(22))/(2)=3xx10^(22)` |
|
| 2038. |
which is are correct statement for following : `Mn^(+4),Cr^(+2),Ni^(+2),Zn^(+2)`A. `Cr^(+2)` have maxiumum magetic momentB. Only `Zn^(+2)` is damagneticC. `Mn^(+4)` have maximum magnetic momentD. `Ni^(+2)` have maximum magnetic moment |
| Answer» Correct Answer - A,B | |
| 2039. |
The reaction between `H_(2)` and `Br^(2)` to form `HBr` in presence of light is initiated by the photo decomposition of `Br_(2)` into free Br atoms (free radicals) by absorption of light. The bond dissociation energy of `Br_(2)` is `192kJ//"mole"`.What is the longest wavelength of the photon that would initiate the reaction? |
|
Answer» Correct Answer - `6235A^(@)` `H_(2)+Br_(2)overset(hv)rarr2HBr` `Br_(2)overset(hv)rarr2Br` `BE=192kJ//"mole"` `(192)/(93.368)eV//"mole"=(hc)/(hv)` or `(192)/(96.368)=(1240)/(lamda(nm))` `lambda=6235Å` |
|
| 2040. |
Calculate the wavelength of a rested electron (in "Å") after it absorbs a photon of wavelength 9 nm. `["Given"h=6xx10^(-34) J-sec,m=9xx10^(-31) kg].` |
|
Answer» Correct Answer - 1 |
|
| 2041. |
Which of the following electron transition in hydrogen atom will require the energy equivalent to its ionization energy?A. from `n = 1` to `n = 2`B. from `n = 2` to `n = 3`C. from `n = 1` to `n = 3`D. from `n = 1` to `n = oo` |
|
Answer» Correct Answer - D In ionisation `e^(-)` jump from n orbit to `oo` orbit |
|
| 2042. |
For hydrogen atom, radius of first Bohr orbit would beA. `9.63xx10^(-10) m`B. `0.43xx10^(-10)m`C. `0.50xx10^(-10)m`D. `0.53xx10^(-10)m` |
|
Answer» Correct Answer - D `r=(n^(2)h^(2))/(4pi^(2)me^(2)z)` `n=1, h=6.62xx10^(-34)Js^(-1)` `:.r=0.53xx10^(-10)m=0.53 Å` |
|
| 2043. |
The following electonic transition corresponds to the shortest wave length (n= no. of orbit)A. `n_(5) rarr n_(1)`B. `n_(5) rarr n_(3)`C. `n_(5) rarr n_(2)`D. `n_(5) rarr n_(4)` |
|
Answer» Correct Answer - A `lambda prop (1)/(Deltan)` |
|
| 2044. |
How many electron in an atom with atomic number `105` can have `(n + l) = 8` ?A. ` 30`B. `17`C. ` 15`D. unpredictable |
|
Answer» Correct Answer - B Electronic configuration of atom with atom with atomic no. 105 is `: 1s^2 . 2s^2 . 2s^2 2s^2 2p^6, 3s^2 3p^6 3d^(10) , 4d^(10), 4s^6 4p^6 4d^(10)` ` 4f^(14), 5s^2 5p^6 5d^(10) underline(5g^(14)), 6s^2 6p^6 underline (6d^3) , 7s^2` The underline orbitals (5 f and 6d) have ( n +l) =8. |
|
| 2045. |
(i) Calculate the number of electrons which will together with one gram . (ii) Calculate the mass and charge on one mole of electrons . |
|
Answer» `(i) 9. 108 xx 10^(28) g` = 1 electron ` therefore 1 g= 1/( 9. 108 xx 10^(-28))` ` = 1.097 xx 10^(27)` elecron (ii) Mass fo (1) mole electron ` = 9. 10 8 xx 10^(-28) xx 6. 0 223 xx 10^(23)` `= 5 . 48 xx 10^(-4) g` Charge of `1` mole electron ` = 1 . 66 xx 10^(-19) xx 6. 023 xx 10^(23)` `= 9. 99 xx 10^4 C`, |
|
| 2046. |
The spin-only magnetic moment [in units of Bohr magneton, `(mu_(B)` of `Ni^(2+))` in aqueous solution would be (atomic number of `Ni=28)`A. `2.84`B. `4.9`C. 0D. `1.73` |
|
Answer» Correct Answer - A `mu_(B) = sqrt(n(n+2))Ni^(+2) = 3d^(8)` Number of unpaired electrons = 2 |
|
| 2047. |
`A` golf ball has a mass of `40g` and a speed of `45m//s`. If the speed can be measured within accuracy of `2%`, calculate the uncertainty in the position. |
|
Answer» The uncertainty in the speed is 2%, i.e., `45 xx (2)/(100) = 0.9 ms^(-1)` using the equation `Delta x = (h)/(4pi m Delta v)` `= (6.626 xx 10^(-34)Js)/(4 xx 3.14 xx 40 xx 10^(-3)kg xx 0.9 ms^(-1)) = 1.46 xx 10^(-33)m` |
|
| 2048. |
The spin magnetic moment of cobalt in the compound `Hg[Co(SCN)_(4)]` isA. `sqrt(3)`B. `sqrt(8)`C. `sqrt(15)`D. `sqrt(24)` |
| Answer» Correct Answer - C | |
| 2049. |
In collection of H-atom all electrons jump from 5th excited state to ground level finally , without emitting any line in infrared region . Total number of possible different radiations are :A. 9B. 15C. 6D. 3 |
|
Answer» Correct Answer - D |
|
| 2050. |
Imagine an atom made of a proton and a hypothetical particle of double the mass of the electron but having the same change as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle of the first excited level. the longest wavelength photon that will be emitted has wavelength [given in terms of the Rydberg constant `R` for the hydrogen atom] equal toA. `(9)/(5R)`B. `(36)/(5R)`C. `(18)/(5R)`D. `(4)/(R)` |
|
Answer» Correct Answer - C Energy is related to mass: `E_(n)propm` The longest wavelength `lamda_(max)` photon will correspond to the transition of particle from =3 to n=2. `(1)/(lamda_(max))=2R((1)/(2^(2))-(1)/(3^(2)))` `lamda_(max)=(18)/(5R)` |
|