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.
| 451. |
Find the effective mass of a photon if the wavelength of radiation is `3000Å`. |
|
Answer» Correct Answer - `7.367 xx 10^(-36) kg` |
|
| 452. |
Find the effectiveness of a photon of energy 5 eV. |
|
Answer» Correct Answer - `8.889 xx 10^(-36) kg` |
|
| 453. |
Calculate the frequency of the photon, which can excite the electron to -3.4 eV from -13.6 eV. |
|
Answer» Correct Answer - `2.47 xx 10^(15) Hz` |
|
| 454. |
The ground state energy of hydrogen atom is `-13.6eV`. If an electron makes a transition form an energy level `-0.85 eV` to `-3.4 eV`, calculate the wavelength of spectral line emitted. To which series of hydrogen spectrum does this wavelength belongs? |
|
Answer» Correct Answer - 4853 Å, Balmer series |
|
| 455. |
If `13.6eV` energy is required to separate a hydrogen atom into a proton and an electron, then the orbital radius of electron in a hydrogen atom isA. `5.3xx10^(-11)m`B. `4.3xx10^(-11)m`C. `6.3xx10^(-11)m`D. `7.3xx10^(-11)m` |
|
Answer» Correct Answer - A Here, `E=-13.6eV=-13.6xx1.6xx10^(-19)=-2.2xx10^(-18)J` `E=(-e^2)/(8piepsilon_0r)` `therefore` As orbital radius, `r=(-e^2)/(8piepsilon_0E)=(9xx10^(9)xx(1.6xx10^(-19))^2)/(2xx(2.2xx10^(-18)))=5.3xx10^(-11)m`. |
|
| 456. |
It is found experimentally that `13.6 eV` energy is required to separated a hydrogen atom into a proton and an electron. Compute the orbital radius and velocity of electron in a hydrogen atom. |
|
Answer» Total energy of the electron in hydrogen atom is - 13.6 eV = `13.6 xx1.6 xx10^(-19)J=-22xx10^(-18)J` Thus from Eq. (12.4), we have `-(e^(2))/(8piepsilon_(0)r)=-2.2 xx 10^(-18) J` This gives the orbital radius `r=-(e^(2))/(8piepsilon_(0)E)=-((9xx10^(9)Nm^(2)//C^(2))(1.6 xx 10^(-19)C)^(2))/((2)(-2.2xx10^(-18)J))` = `5.3 xx 10^(-11)m`. The velocity of the revolving electron can be computed from Eq. (12.3) with m = `9.1 xx 10^(-31) kg`, `upsilon=(e)/(sqrt(4piepsilon_(0)mr))=2.2 xx 10^(6)m//s`. |
|
| 457. |
In the original experiment, Geiger and Marsden calculated the distance of closest approach to the gold nucleus (Z=79)- of a 7.7MeV `alpha` particle before it comes momentarily to rest and reverses its direction. What is its value? |
|
Answer» The key idea here is that throughout the scattering process,the total mechanical energy of the system consisting of an `alpha`-particle and a gold nucleus is conserved. The system’s initial mechanical energy is `E_(i)`, before the particle and nucleus interact, and it is equal to its mechanical energy `E_(f)` when the `alpha`-particle momentarily stops. The initial energy `E_(i)` is just the kinetic energy K of the incoming - particle. The final energy `E_(f)` is just the electric potential energy U of the system. The potential energy U can be calculated from Eq. (12.1). Let d be the centre-to-centre distance between the `alpha`-particle and the gold nucleus when the `alpha`-particle is at its stopping point. Then we can write the conservation of energy `E_(i) = E_(f)` as `K=(1)/(4piepsilon_(0))((2e)(Ze))/(d)=(2Ze^(2))/(4piepsilon_(0)d)` Thus the distance of closest approach d is given by `d=(2Ze^(2))/(4piepsilon_(0)K)` The maximum kinetic energy found in `alpha`-particles of natural origin is 7.7 MeV or `1.2 xx 10^(12)`J. Since `1//4piepsilon_(0)=9.0xx10^(9)Nm^(2)//C^(2)`. Therefore with e = `1.6 xx 10^(-19)C`, we have, `d=((2)(9.0xx10^(9)Nm^(2)//C^(2))(1.6xx10^(-19)C)^(2)Z)/(1.2 xx 10^(-12)J)` `=3.84 xx 10^(-16) Zm` The atomic number of foil material gold is Z = 79, so that d (Au) = `3.0 xx 10^(-14)m` = 30 fm. (1 fm (i.e. fermi) = `10^(-15)` m.) The radius of gold nucleus is, therefore, less than `3.0 × 10^(–14)` m. Thisis not in very good agreement with the observed result as the actual radius of gold nucleus is 6 fm. The cause of discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the `alpha`-particle. Thus, the `alpha`-particle reverses its motion without ever actually touching the gold nucleus. |
|
| 458. |
In a geiger - marsden experiment. Find the distance of closest approach to the nucleus of a 7.7 me v `alpha`- particle before it comes momentarily to rest and reverses its direction. (z for gold nucleus = 79) .A. 10 fmB. 20 fmC. 40 fmD. 10 fm |
|
Answer» Correct Answer - C Let d be the distance of closest approach then by the conservation of energy. Initial kinetic energy of incoming `alpha`-particle K. = Final electric potential energy U of the sysyem As `K=(1)/(4piepsilon_0)xx((2e)(Ze))/(d)therefored=(1)/(4piepsilon_0)(2Ze^2)/(K)`....(i) Here, `(1)/(4piepsilon_0)=9xx10^9Nm^2C^(-2)`,`Z=79,e=1.6xx10^(-19)C`. `K =7.7MeV=7.7xx10^6xx1.6xx10^(-19)J=1.2xx10^(-12)J` Substituting these values in (i) `d= (2xx9xx10^(9)xx(1.6xx10^(-19))^2xx79)/(1.2xx10^(-12))` `d=3xx10^(-14)m=30fm` (`because 1 fm=10^(-15)m`) |
|
| 459. |
Name the constituents of an atomic nucleus. What is a nucleon? |
|
Answer» The constituents of an atomic nucleus are (1) the proton, a positively charged particle (2) the neutron, a neutral (uncharged) particle. The term nucleon (nuclear constituent) refers to a proton as well as a neutron. [Note : The electron was discovered by J. J. Thomson in 1897. The proton was discovered by Ernest Rutherford in 1919. The neutron was discovered in 1932 by James Chadwick (1891- 1974), British physicist. The existence of the neutron and the deuteron was predicted by Rutherford in 1920. The proton has a mass about 1836 times that of the electron, but the magnitude of the electric charge is the same for both. The mass of the neutron is slightly more than that of the proton.] |
|
| 460. |
What is a nuclear reactor? |
|
Answer» A nuclear reactor is a device in which a nuclear fission chain reaction is used in a controlled manner (i) to produce energy in form of heat which is then converted into electricity or (ii) to produce radioisotopes or (iii) to produce new nuclides using a suitable fissionable material such as uranium or plutonium. In a uranium reactor, \(^{235}_{92}U\) is bombarded by slow neutrons to produce \(^{235}_{92}U\) which undergoes fission. |
|
| 461. |
What is the emission spectrum of a substance? Explain in brief. |
|
Answer» The emission spectrum of a substance is the distribution of electromagnetic radiations emitted by the substance when it is heated, or bombarded by electrons, or ions or photons. The distribution is arranged in order of increasing (or decreasing) frequency (or wavelength) and is characteristic of the substance unless the temperature of the substance is very high when the distribution is continuous. The spectrum may be a line spectrum or band spectrum. The intensities corresponding to different frequencies are different. [Note : The absorption spectrum is formed by absorption of electromagnetic radiation when the substance is exposed to radiation of all frequencies.] |
|
| 462. |
What is the shortest wavelength present in the Paschen series of spectral lines?A. 720 nmB. 790 nmC. 800 nmD. 820 nm |
|
Answer» Correct Answer - D The wavelength for Paschen series, `(1)/(lambda)=R[(1)/(3^2)-(1)/(n^2)]` For shortest wavelength `n=prop` `therefore (1)/(lambda)=R[(1)/(9)-(1)/(prop^2)]=(R)/(9)rArrlambda=(9)/(R)=(9)/(1.097xx10^7)` `=8.20xx10^-7m=820nm`. |
|
| 463. |
State the observations which lead to the conclusion that radioactivity is a nuclear phenomenon. |
|
Answer» The rate of disintegration of a radioactive material is not affected by changes in physical and chemical conditions such as (1) temperature and pressure (2) action of electric and magnetic fields (3) chemical composition of the material. The above changes affect the orbital electrons, but not the nucleus. Therefore, we conclude that radioactivity is a nuclear phenomenon. |
|
| 464. |
Who discovered radioactivity? Give a brief account of the first observation of radioactivity. |
|
Answer» Antoine Henri Becquerel (1852-1908), French physicist, discovered radioactivity in 1896. He had kept photographic plates wrapped in a thick black paper in a drawer of his desk. Later, he also kept uranium salts near the photographic plates. After some days he developed the photographic plates and was surprised to find that they were fogged although he had protected them from light. Becquerel concluded that uranium salts must be emitting some invisible rays which affected the photographic plates. In this way, radioactivity was discovered by Becquerel. |
|