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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.
| 101. |
The short wavelength limits of Lyman, Paschen and Balmer series in the hydrogen spectrum are denoted by `lambda_L, lambda_P and lambda_B` respectively. Arrange these wavelength in increasing order. |
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Answer» As `1/lambda prop(1/(n_1^2)-1/(n_2^2))` `1/lambda prop 1/(n_1^2) (because n_2=oo)` for Lyman series `n_1=1` for Balmer series `n_1=2` for Paschen series `n_1=3`. `:. lambda_(L) lt lambda_(B) lt lambda_(P)`. |
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| 102. |
1curei=k disintegrations/sec, where k isA. `3.7xx10^(10)`B. `3.7xx10^(-10)`C. `7.3xx10^(-10)`D. `7.3xx10^(10)` |
| Answer» Correct Answer - A | |
| 103. |
Which of the following is in the increasing order for penetrating power ?A. `gamma, beta, alpha`B. `gamma, alpha, beta`C. `alpha, beta,gamma`D. `alpha,gamma, beta` |
| Answer» Correct Answer - C | |
| 104. |
Which of the following is in the increasing order for penetrating power ? |
| Answer» `alpha,beta and gamma`-rays | |
| 105. |
During a nuclear fusion reaction,A. A heavy nucleus breaks into two fragments by itselfB. a light nucleus bombarded by thermal neutrons breaks upC. A heavy nucleus bombarded by thermal neutrons breaks upD. Two light nuclei combine to give a heavier nucleus and possibly other products |
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Answer» Correct Answer - D Nuclear fusion is the process in which two or more light nuclei combine to give a heavy nucleus with the emission of log of energy. |
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| 106. |
Which of the following statements (s) is (are) correct?A. The rest mass of a stable nucleus is less than the sum of the rest masses of its separated nucleonsB. The rest mass of a stable nucleus is greater than the rest masses of its separated nucleonsC. The nuclear fusion, energy is released by fusing two nuclei of medium mass. (approx. 100 am u)D. In nuclear fission, energy is released by fragmentation of a very heavy nucleus. |
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Answer» Correct Answer - A::D As some mass is converted into binding energy of nucleus, so rest mass of nucleons is greater than the mass of stable nucleus and in fission reaction, a heavy nucleus breaks down into smaller nuclei with the emission of energy. |
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| 107. |
As the mass number `A` increases, which of the following quantities related to a nucleus do not change?A. massB. volumeC. densityD. binding energy |
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Answer» Correct Answer - C As mass number A increases, mass, volume and binding energy all increases, but density remains constant. |
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| 108. |
A source contains two phosphorus radionuclides `._(15)P^(35) (T_(1//2)=14.3"days")` and `._(15)P^(33) (T_(1//2)=25.3"days")`. Initially, `10%` of the decays come form `._(15)P^(35)`. How long one must wait until `90%` do so? |
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Answer» Suppose, initially the source has `90% ._(15)P_(1)^(32)` and `10% ._(15)P_(2)^(32)`, say 9x gram of `P-2` and x gram of `P_1`. After t days, suppose the source has `90% ._(15)P_(2)^(33)` and `10% ._(15)P_(1)^(32)` i.e, y gram of `P_2` and 9y gram of `P_1`. we have to calculate t. form `N/(N_0)=(1/2)^n=(1/2)^(t//T)=2^(-t//T)` `N=N_(0) 2^(-t//T) :. y=(9x) 2^(-t//14.3)`, for P-2 and `(9y)=x 2^(-t//25.3)`, for `P_1` Dividing, we get `1/9=9xx2^((t//25.3-t//14.3))` or `1/81=2^(-11t//25.3xx14.3)` `log 1-log 81=(-11t)/(25.3xx14.3) log 2` `0-1.9085=(11t)/(25.3xx14.3)xx0.3010` `t=(25.3xx14.3xx1.9085)/(11xx0.3010)=208.5 days` |
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| 109. |
The selling rate of a radioactive isotope is decided by its activity. What will be the second-hand rate of a oe month old `P^32``(t_(1//2)=14.3` days) source if it was originally purchased for 800 rupees? |
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Answer» Correct Answer - `Rs.182.8` Here, `T=14.3days, t=30 days` `lambda=0.693/T=0.693/14.3day^(-1)` As `A=A_(0)e^(-lambdat)` `:. A/(A_(0))=e^((-30xx0.693)/14.3)=e^(-1.454)` `log_(e).(A)/(A_(0))=-1.454 log_(e) e=-1.454` `2.303 log_(10) .(A)/(A_(0))=-1.454` `log_(10). (A)/(A_(0))=-1.454/2.303=-0.631=bar(1).369` `A/(A_(0))=0.2285` `:.` Sal e price =`0.2285xx800` `=182.8 rupees` |
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| 110. |
`gamma`-rays are also called electromagnetic waves. Why? |
| Answer» `gamma`-rays are photons of short wavelength. They travel with the speed of em waves and exhibit those phenomena which are being exhibated by electromagnetic waves. Hence `gamma`-rays are called electromagnetic waves. | |
| 111. |
A nucleus contains no electrons, yet it ejects them. How? |
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Answer» A neutron inside a nucleus is unstable. It decays into a proton, an electron and an antineutrino. `._0n^1` to `._1H^1+._-1e^0+barv` It is this electron, which is emitted as a `beta`particle form the nucleus. |
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| 112. |
A radioactive sample having N nuclei has activity R. Write down an expression for its half life in terms of R and N. |
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Answer» Activity , `R=lambdaN :. lambda=R//N` Half life, `T=0.693/lambda=(0.693N)/R` |
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| 113. |
Calculate packing fraction of alpha particle form the following date: `m_(alpha)=4.0028a.m.u., m_(p)=1.00758a.m.u. , and m_(n)=1.00897 a.m.u.` |
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Answer» Correct Answer - `7.58xx10^(-3)a.m.u.//N` An alpha particle contains two protons and two neutrons. Mass excess, `Deltam=2m_(p)+2m_(n)-m_(alpha)` `=2xx1.00785+2xx1.00897-4.0028` `=0.0303 a.m.u` Packing fraction `=(Deltam)/A=0.0303/4` `=7.58xx10^(-3)a.m.u//N` |
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| 114. |
The activity of a radioactive sample falls from `600 s^(-1) to 500 s^(-1) in 40 minutes. Calculate its half-life. |
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Answer» Correct Answer - `152mi n` form `A=A_(0)e^(-lambdat)` `500=600e^(-lambdat)` `e^(lambdat)=600/500` `lambdat log_(e) e=log_(e) (6//5)` `lambda=(log_(e) (6//5))/t` Half life, `T=0.693/(lamdba)=(0.693xxt)/(log_(e) (6//5))` `T=(0.693xx40)/(2.303xx0.792)=152min` |
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| 115. |
What are unit of activity of radioactive elements? |
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Answer» 1 curie=`3.7xx10^(10)` disintegration/sec, 1 rutherford =`10^6` disintegration/sec and 1bacquerel =1Bq=1decay/sec. |
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| 116. |
The nucleus `.^(23)Ne` deacays by `beta`-emission into the nucleus `.^(23)Na`. Write down the `beta`-decay equation and determine the maximum kinetic energy of the electrons emitted. Given,`(m(._(11)^(23)Ne) =22.994466 am u` and `m (._(11)^(23)Na =22.989770 am u`. Ignore the mass of antineuttino `(bar(v))`. |
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Answer» The `beta^(-)` decay of `._10Ne^(23)` may be represented as `._10Ne^(23)to ._11Na^(23) +._(-1)e^0+barv+Q` Ignoring the rest mass antineutrino `(barv)` and electron Mass defect, `Deltam=m(._10Na^(23))-m(._11Na^(23)) =22.994466-22.989770-0.004696a.m.u` `:. Q= 0.004696xx931MeV=4.372MeV` As `._11Na^(23)` is very massive, this energy of `4.3792MeV`, is shared by `e^(-) and barv` pair. The max. K.E. of `e^(-)=4.372 MeV`, when energy carried by `barv` is zero. |
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| 117. |
Why is neutron so effective as a bombarding particle? |
| Answer» This is primarily because neutrons carries no charge.It is neither attracted nor repelled by nucleus and electrons. Therefore, neutron is best projectile. | |
| 118. |
Calculate the disintegration energy Q for fission of `._42Mo^(98)` into two equal fragments `._21Sc^(49)` by bombarding with a neutron. Given that `m(._42Mo^(98))=97.90541u, m(._21Sc^(49))=48.95002u, m_n=1.00867u` |
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Answer» The disintegration energy in fission of `._42Mo^(82)` is given by `Q=(Deltam)xx931MeV=[m(._42Mo^(82))+m_n-2m(._21Sc^(49))]xx931MeV` `=[97.90541+1.00867-2xx48.95002]xx931MeV` `Q=1.01404xx931=944.1MeV` |
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| 119. |
A star converts all its hydrogen to helium, achieving 100% helium composition. It then converts the helium to carbon via the reaction `._2He^4 + ._2He^4+ ._2He^4 to._6C^(12)+7.27MeV` The mass of the star is `5.0xx10^(32)kg` and it generates energy at the rate of `5xx10^(30)kg` watt. How long will it take to convert all the helium to carbon at this rate? |
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Answer» Mass of star `=5.0xx10^(32)kg=5.0xx10^(35)g` Number of atoms in 4grams of He `=6.023xx10^(23)` `:.` Number of atoms in `5.0xx10^(35)` gram of He `=(6.023xx10^(23)xx5.0xx10^(35))/4=7.529xx10^(58)` As each fusion reaction consumed 3 helium atoms to produce 7.27MeV energy. `:.` Total energy produced by fusion of `5.0xx10^(35)` gram of He `7.27/3xx7.529xx10^(58)MeV` `=(7.27xx7.529xx10^(58)xx1.6xx10^(-13))/3J` `=29.191473xx10^(45)J` As power `=5xx10^(30)"watt" =5xx10^(30)"joule"//sec`. `:.` Time required `= (29.191473xx10^(45))/(5xx10^(30))sec`. `=(29.191473xx10^(15))/(5xx60xx60xx24xx365) years` `t=1.85xx10^8 years` |
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| 120. |
Statement-1: After ten half life lives, the amount of a radioactive element reduces to about `1//1000` part. Statement-2: `N=N_(0)(1/2)^(n)=N_(0)(1/2)^(10)` `=(N_(0))/1024 ~~(N_(0))/1000`A. Statement-1 is true, Statement-2 is true and Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, but Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, but Statement-2 is falseD. Statement-1 is false, but Statement-2 is true. |
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Answer» Correct Answer - A Both the statement are true and statement-2 is correct explanation of the statement-1. |
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| 121. |
Which law is violated in the following nuclear reaction? `._0n^1 to ._1H^1+._(-1)e^0`? |
| Answer» The law of conservation of spin (angular momentum) is violeted. Each particle on right side has a spin `1/2(h//2pi)` so that the resultant spin on the right side is 0 or `1(h//2pi)`. And on the left side, is `1/2(h//2pi)`. | |
| 122. |
`M_(1) and M_(2)` represent the masses of `._(10)Ne^(20)` nucleus and `._(20)Ca^(40)` nucleus respectively. State whether `M_(2)=2M_(1)` or `M_(2)gt2M_(1)` or `M_(2)lt2M_(1)` |
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Answer» Correct Answer - `M_(2)>2M_(1)` Let `m_(p)` be the mass of a proton and `m_(n)` be the mass of a neutron. As some energy is required to bind a nucleus, the mass of the nucleus is always less than the mass of constituent nucleons. Therefore, `M_(1)lt10(m_(p)+m_(n))` `M_(2)lt20(m_(p)+m_(n))` As BE/nucleon of `._(20)Ca^(40)` is more than that of `._(10)Ne^(20)`, the factor `(M_(2)-20(m_(p)+m_(n)))/40gt (M_(1)-10(m_(p)+m_(n)))/20` or `M_(2)-20(m_(p)+m_(n))gt 40/20[M_(1)-10(m_(p)+m_(n))]` of `M_(2)gt2M_(1)` |
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| 123. |
The half life of a radioactive element A is same as mean life time of another radioactive element B.Initially, both have same number of atoms. B decays faster than A . Why? |
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Answer» `T_A=tau_B=1.44 T_B :. T_AgtT_B` `lambda_A lt lambda_B`. Therefore, B decays faster than A. |
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| 124. |
(a) Write the basic nuclear process involved in the emission of `beta^(+)` in a symbolic form, by a radioactive nucleus. (b) In the reaction given below: `._6C^(11) to ._yB^z=x+v` `._6C^(12)+._6C^(12) to ._aNe^(20)+._bHe^c` Find the values of x,y,z and a,b,c. |
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Answer» (a) The basic nuclear process involved in beta plus decay is transformation of a proton into a neutron within the nucleus according to `._1p^1to._0n^1+e^(+)+v("neutrino")` (b) Applying the principle of conservation of charge number and mass number in the given nuclear reactions, we get (i) `._6C^(11)to._yB^z+x+v` x is an positron, v is neutrino Now, `11=z+0+0 , z=11` `6=y+1+0,y=5` (ii) `._6C^(12)+._6C^(12)to._aNe^(20)+._bHe^c` He is `._2He^4 :. b=2,c=4` Again, `12+12=20+c :. c=4` Also, `6+6=a+b=a+2, :. a=10` |
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| 125. |
The half-life period of a radioactive element x is same as the mean life time of another radioactive element y. Initially, both of them have the same number of atoms. Then, (a) x and y have the same decay rate initially (b) x and y decay at the same rate always (c) y will decay at a faster rate than x (d) x will decay at a faster rate than yA. X and Y have the same decay rate intiallyB. X and Y decay at the same rate alwaysC. Y will decay at a faster rate than XD. X will decay at a faster rate than Y |
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Answer» Correct Answer - C As given `(T)_(x)=tau_(y) or 0.693/(lambda_(x))=1/(lambda_(y))` `lambda_(x)=0.693lambda_(y) or lambda_(x)gtlambda_(x)` As rate of decay =`lambdaN` So Y decays at faster rate than X. |
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| 126. |
The half life of a freshly prepared radioactive sample is 2h. If the sample emits radiation of intensity, which is 16 times the permissible safe level, then the minimum time taken after which it would be possible to work safely with source is: |
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Answer» Correct Answer - 8 Here, `T=2h, t=?` To work safely with the sample, its activity must be reduced to `1/16`. form `N/(N_(0))=(1/2)^(n)=1/16=(1/2)^(4)` `:. n=4` `t=nT=2xx4=8h` |
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| 127. |
A freshly prepared radioactive source of half-life `2 h` emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source isA. `6h`B. `12h`C. `24h`D. `28h` |
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Answer» Correct Answer - B As `(N_(0))/N=2^(t//T),. (N_(0))/N=64/1=2^(t//T)` `2^(6)=2^(t//T), t=6T` `t=6xx2=12h` |
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| 128. |
Two deutrons each of mass m fuse to form helium resulting in release of energy E. The mass of helium formed isA. `m+E//c^(2)`B. `2m+E//c^(2)`C. `E//mc^(2)`D. `2m-E//c^(2)` |
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Answer» Correct Answer - D Total mass before fusion =2m Mass converted into energy in this process `=E//c^(2)` `:.` Mass of He formed =`(2m=E//c^(2))` |
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| 129. |
What is the ground state energy of electron in case of `Li^(2+)` ?A. `13.6eV`B. `-13.6eV`C. `30.4eV`D. `-30.4eV` |
| Answer» Correct Answer - D | |
| 130. |
What is the order of velocity of electron in a hydrogen atom in ground state?A. `10m//s`B. `10^(6)m//s`C. `10^(-6)m//s`D. `10^(7)m//s` |
| Answer» Correct Answer - B | |
| 131. |
What is the order of velocity of electron in a hydrogen atom in ground state? |
| Answer» Correct Answer - `10^6ms^-1` | |
| 132. |
Orbital velocity of electron is outer orbits is............as compared to its value in............. . |
| Answer» Correct Answer - less ; inner orbits, | |
| 133. |
The ratio of radii of orbits correponding to first and second excited states of hydrogen atom isA. `1`B. `1:2`C. `2:3`D. `4:9` |
| Answer» Correct Answer - D | |
| 134. |
Name the series of hydrogen atom which lies in U.V. region. |
| Answer» Lyman series is in U.V. region. | |
| 135. |
Total energy of electron in outer orbits is........that in............. . |
| Answer» Correct Answer - more than ; inner orbits. | |
| 136. |
What is the ratio of radii of orbits corresponding to first excited state and ground state in hydrogen atom? |
| Answer» `(r_2)/(r_1)=(n_2^2)/(n_1^2)=(2^2)/(1^2)=4:1` | |
| 137. |
Define energy level. How is it represented? |
| Answer» Energy level is represented by a horizontal line drawn according to some suitable energy scale which represents the total energy of electron in a stationary orbit of an atom. | |
| 138. |
What are the values of first and second excitation potential of hydrogen atom? |
| Answer» Correct Answer - 10.2V , 12.09V. | |
| 139. |
What is the ionisation potential of hydrogen atom?A. `-13.6eV`B. `13.6eV`C. `-13.6eV`D. `13.6eV` |
| Answer» Correct Answer - D | |
| 140. |
The wavelength of the first line of Balmer series in hydrogen atom is `6562.8Å`. Calculate ionisation potential of hydrogen and also, the wavelength of first line of Lyman series. |
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Answer» Here, `lambda_1=6562.8Å=6562.8xx10^(-10)m` `:. 1/(lambda_1)=R(1/(2^2)-1/(3^2))=Rxx5/36` `R=36/(5lambda_1)=36/(5xx6562.8xx10^(-10))` `R=1.0971xx10^7m^-1` The energy of electron in nth energy state `E_n=R(hc)/(n^2)` `:.` Ionisation energy of Hydrongen atom `E=E_1-E_(oo)=Rhc-0` `=1.0971xx10^7xx6.6xx10^(-34)xx3xx10^8` `=21.788xx10^(-19)J` `=(21.78xx10^(-19))/(1.6xx10^(-19))eV=13.6eV` `:.` Ionisation potential of hydrogen atom `=13.6V` The wavelength of first line of Lyman series is given by `1/lambda=R(1/(1^2)-1/(2^2))=3/4R` `lambda=4/(3R)=4/(3xx1.0971xx10^7)` `=1.2153xx10^-7m=1215.3Å` |
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| 141. |
What is the ionisation potential of hydrogen atom? |
| Answer» Correct Answer - 13.6V | |
| 142. |
Write an empirical relation for Paschen series of hydrogen spectrum. |
| Answer» `barv=R(1/(3^2)-1/(n^2))`, where n=4,5,6.... | |
| 143. |
Calculate longest wavelength of Paschen series. Given `R=1.097xx10^(7)m^(-1)`. |
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Answer» Correct Answer - `18752A^(@)` Take `n_(1)=3 and n_(2)=4` |
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| 144. |
The seond line of Balmer series has wavelength `4861 Å` The wavelength o fthe first line Balmer series is |
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Answer» Correct Answer - `6562Å` For `H_(alpha)` line, `n_(1)=2, n_(2)=3` and for `H_(beta)` line `n_(1)=2, n_(2)=4` |
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| 145. |
The total energy of an electron in second excited state of hydrogen atom is -1.51eV. Calculate (i) KE of electron (ii) PE of electron in this state. |
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Answer» As is known, P.E. of electron`=-2KE` of electron. Total energy `=PE+KE=-2KE+KE=-KE` `-1.51=-KE :. Ke=1.51eV` and `PE=-2KE=-2xx1.51eV=-3.02eV` |
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| 146. |
The first four spectral lines in the Lyman series of a H-atom are `lambda=1218Å, 1028Å,974.3Å and 951.4Å`. If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines. |
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Answer» In hydrogen atom, one electron (of mass `m_e`) revolves around one proton (of mass M) Reduced mass for hydrogen, `mu_(H)=(m_exxM)/(m_e+M)=(m_e)/((1+(m_e)/M)` In deuterium, `._1D^2`, one electron ( of mass `m_e`) revolves around nucleus containing one proton and one neutron (of mass 2M). `:.` Reduced mass for deuterium, `mu_(D)=(m_exx2M)/(2M+m_e)=(2M.m_e)/(2M(1+(m_e)/(2M)))=m_e(1-(m_e)/(2M))` When an electron jumps form orbit j to orbit i, the frequency of radiation emitted `hv_(ji)=(E_(j)-E_(i)) propmu` (reduced mass) `:. lambda_(ji) prop 1/mu` If `lambda_(D)` is wavelength emitted in case of deuterium, and `lambda_(H)` is wavelength emitted in case of hydrogen, then ` (lambda_(D))/(lambda_(H))=(mu_(H))/(mu_(D))=(m_e(1-(m_e)/M))/(m_e(1-(m_e)/(2M)))=(1-(m_e)/M)(1-(m_e)/(2M))^(-1)=(1-(m_e)/M)(1+(m_e)/(2M))` `=(1-(m_e)/M+(m_e)/(2M))=(1-(m_e)/(2M))` As `(m_e)/M=1/1840`, therefore `(lambda_(D))/(lambda_(H))=(1-1/(2xx1840))=0.99973` `lambda_(D)=(0.99973)lambda_(H)` Using `lambda_(H)=1218Å,1028Å,974.3Å and 951.4Å`, we get `lambda_(D)=1217.7Å,1027.7Å,974.04Å,951.1Å`. |
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| 147. |
Why is electron supposed to be revolving around the nucleus? |
| Answer» If the electrons were stationary, they would fall into the nucleus due to electrostatic attraction and the atom would be unstable. | |
| 148. |
The K.E. of `alpha`-particle incident gold foil is doubled. How does the distance of closest approach change? |
| Answer» As is known, distance of closest approach varies inversely as K.E. of `alpha`-particle. When K.E. is doubled, `r_0` is halved. | |
| 149. |
In hydrogen atom, wavelength of emitted photon will be minimum in which of the following transitions? (i) `n = 2` to `n = 1` , (ii) `n=6` to `n = 5` , (iii) `n = 3` to `n =2` |
| Answer» Wavelength of emitted photon will be minimum when frequency and energy of photon are maximum. for maximum energy, the energy difference of the two transitions should be maximum. This occurs between n=2 to n=1. | |
| 150. |
Find the wavelength of electron orbiting in the first excited state of hydrogen atom. |
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Answer» Here, `lambda`=? n=2 for first excited state `v=(2piKe^2)/(nh)` `lambda=h/(mv)=(h,nh)/(2pimKe^2)=(nh^2)/(2pimKe^2)` `lambda=(2(6.6xx10^(-34))^(2))/(2xx3.14xx9xx10^(-31)(9xx10^9)(1.6xx10^(-19))^(2))` `lambda=(2xx6.6xx6.6xx10^-8m)/(6.28xx81xx1.6xx1.6)=6.69xx10^(-10)m` |
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