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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.
| 351. |
Does the stopping potential in photoelectric emission depend upon the intensity of the incident radiation in a photo cell? Comment on it. |
| Answer» No, the stopping potential does not depend upon the intensity of incident radiation but depends on the nature of photosenstive surface and frequency of the incident radation. | |
| 352. |
The variation of wavelength `lambda` of the `K_(alpha)` line with atomic number `Z` of the target is shown by the following curve of A. `A`B. `B`C. `C`D. None of these |
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Answer» Correct Answer - C For `K_(alpha)` line `v prop (Z - 1)^(1) rArr lambda prop (1)/((Z - 1)^(2))` |
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| 353. |
The `K_(alpha)` `X` - rays arising from a cobalt `(z = 27)` target have a wavelength of `179 pm`. The `K_(alpha)` `X` - rays arising from a nickel target `(z = 28)` isA. `gt 179 pm`B. `lt 179 pm`C. `= 179 pm`D. None of these |
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Answer» Correct Answer - B `lambda_(k_(a)) prop (1)/((Z - 1)^(2)) rArr (lambda_(Ni))/(lambda_(Co)) = ((Z_(Co) - 1)/(Z_(Ni) - 1))^(2) = (( 27 - 1)/(28 - 1))^(2)` `l_(Ni) = ((26)/(27))^(2) xx lambda_(Co) = ((26)/(27))^(2) xx 179` `= 165.9 pm lt 179 pm`. |
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| 354. |
A laser beam is used for carrying our surgery because itA. is highly monochromaticB. is highly coherentC. can be sharply focussedD. is highly directional |
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Answer» Correct Answer - C LASER is the short form of light Amplification by Stimultaded Emission of Radiation. It is a device by which an intense, monochromatic,collimated and highly coherent light beam can be obtained . Laser beam is used in surgery because it can be sharply focussed. |
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| 355. |
Photoelectric emission is observed from a metallic surface for frequencies `v_(1) and v_(2)` of the incident light. If the maximum value of kinetic energies of the photoelectrons emitted in the two cases are in the ratio `n:1` then the threshold frequency of the metallic surface isA. `(v_(2)-v_(1))/(n-1)`B. `(nv_(2)-v_(1))/(n-1)`C. `(nv_(1)-v_(2))/(n-1)`D. `(v_(2)-v_(1))/(n)` |
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Answer» Correct Answer - b `(E_(1))/(E_(2))=(h(v_(1)-v_(0)))/(h(v_(2)-v_(0)))=n/1` On solving, `v_(0)=(nv_(2)-v_(1))/(n-1)` |
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| 356. |
Photoelectric emission occurs only when the incident light has more than a certain minimumA. wavelengthB. frequencyC. amplitudeD. angle of incidence |
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Answer» Correct Answer - B In photoelectric effect for a given photosensitive material , there exists a certain minimum cut - off frequency , called the threshold frequency , below which no emission of photoelectrons takes place no matter how intense the light is. |
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| 357. |
The ratio of moment of an electron and an `alpha`-particle which are accelerated from rest by a potential difference of `100V` is |
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Answer» Momentum p=mv and `v=sqrt((2QV)/m)` `p=sqrt(2QmV)` `rArr p prop sqrt(Qm)` `rArr p_e/p_alpha=sqrt((exxm_e)/(2exxm_alpha))=sqrt(m_e/(2m_alpha)) ` |
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| 358. |
The energy that should be added to an electron, to reduce its de-Broglie wavelengths from `10^(-10) m` to `0.5 xx 10^(-10)` m wil be |
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Answer» `lambda=h/sqrt(2mE)` `rArr lambda prop 1/sqrtE` `lambda_1/lambda_2=sqrt(E_2/E_1)` `rArr 10^(-10)/(0.5xx10^(-10))=sqrt(E_1/E_2)` `rArr E_2=4E_1 ` hence , added energy =`E_2-E_1=3E_1` |
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| 359. |
If the de - Broglie wavelengths for a proton and for a `alpha` - particle are equal , then the ratio of their velocities will beA. `4 : 1`B. `2 : 1`C. `1 : 2`D. `1 : 4` |
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Answer» Correct Answer - A As `lambda = (n)/(mv) rArr lambda (1)/(m)` Hence , `(lambda p)/(lambda alpha) = (m_(alpha) v_(alpha))/(m_(p) v_(p))/(m_(p) v_(p)) rArr (1)/(1) = ( m_(alpha) v_(alpha))/(m_(p) v_(p)) rArr (v_(p))/( v_(alpha)) = (m_(alpha))/(m_(p)) = (4)/(1)` |
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| 360. |
Find the ratio of velocities of proton and `alpha`-particle if the de Broglie wavelengths of both the particles is same. |
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Answer» `lambda=h/(m_1v_1)=h/(m_2v_2)` `therefore v_1/v_2=m_2/m_1=4/1` |
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| 361. |
If alpha particle, proton and electron move with the same momentum, them their respective de Broglie wavelengths `lamda_(alpha),lamda_(p),lamda_(e)` are related asA. `lamda_(alpha)=lamda_(p)=lamda_(e)`B. `lamda_(alpha)ltlamda_(p)ltlamda_(e)`C. `lamda_(alpha)gtlamda_(p)gtlamda_(e)`D. `lamda_(p)gtlamda_(e)gtlamda_(alpha)` |
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Answer» Correct Answer - A (a) : de Broglie wavelength , `lamda=(h)/(p)` where symbols have their usual meaning. `becausep_(alpha)=p_(p)=p_(e)" "becauselamda_(alpha)=lamda_(p)=lamda_(e)` |
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| 362. |
A proton, a neutron, an electron and an `alpha`-particle have same energy. Then their de-Broglie wavelengths compare asA. `lamda_(p)=lamda_(n)gtlamda_(e)gtlamda_(alpha)`B. `lamda_(alpha)ltlamda_(p)=lamda_(n)ltlamda_(e)`C. `lamda_(e)ltlamda_(p)=lamda_(n)gtlamda_(alpha)`D. `lamda_(e)=lamda_(p)=lamda_(n)=lamda_(alpha)` |
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Answer» Correct Answer - B (b) : Kinetic energy of particle, `K=(1)/(2)mv^(2)` or `mv=sqrt(2mK)` de Bloglie wavelength, `lamda=(h)/(mv)=(h)/(sqrt2mK)` `"For the given value of K",lamdaprop(1)/(sqrt(2mK))` `:.lamda_(p):lamda_(n):lamda_(e):lamda_(alpha)=(1)/(sqrt(m_(p))):(1)/(sqrt(m_(n))):(1)/(sqrt(m_(e))):(1)/(sqrt(m_(alpha)))` `"Since "m_(p)=m_(n),"hence "lamda_(p)=lamda_(n)` `"As "m_(alpha)gtm_(p),"therefore"lamda_(alpha)ltlamda_(p)` `"As " m_(e)ltm_(n),"therefore"lamda_(e)gtlamda_(n)` `"Hence "lamda_(alpha)ltlamda_(p)=lamda_(n)ltlamda_(e)` |
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| 363. |
An electron and a proton are accelerated through the same potential. Which one of the two has (i)b greater value of de-Broglie wavelength associated with it and (ii) less momentum?justify your answer. |
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Answer» When a charged particle of charge q, mass m is accelerated under a potential difference V, Let v be the velocity acquired by particle. Then `qV=1/2mv^2 or mv=sqrt(2qVm)` (i) `lambda=h/(mv)=h/(sqrt(2qVm)) or lambda prop 1/(sqrt(qm))` `:. (lambda_e)/(lambda_p)=sqrt((q_pm_p)/(q_em_e))=sqrt((exx1837m_e)/(exxm_e))gt1` So, `lambda_egtlambda_p`, i.e., greater value of de-broglie wavelength is associated with electron as compared to proton. (ii) Momentum of particle, `p=mv=sqrt(2qVm)` `:. p prop sqrt(qm)`, Hence `(p_e)/(p_p)=sqrt((q_em_e)/(q_pm_p))=sqrt(e/exx(m_e)/(1837m_e)) lt 1` So, `p_e lt p_p`, i.e., lesser momentum is associated with electron as compared to proton. |
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| 364. |
A proton and an alpha - particle are accelerated through same potential difference. Then, the ratio of de-Broglie wavelength of proton and alpha-particle isA. `sqrt(2):1`B. `sqrt(4):1`C. `sqrt(6):1`D. `sqrt(8):1` |
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Answer» Correct Answer - D (d) : As `lamda=(h)/(sqrt(2mqV))` `:.lamdaprop(1)/(sqrt(mq)):.(lamda_(p))/(lamda_(alpha))=(sqrt(m_(alpha)q_(alpha)))/(sqrt(m_(p)q_(p)))` `(sqrt(4m_(p)xx2e))/(sqrt(m_(p)xxe))=sqrt(8)" "( :.m_(alpha)=4m_(p),q_(alpha)=2q_(p))` |
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| 365. |
A proton and an `alpha`-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths `lambda_(p) and lambda_(a)` related to each other? |
| Answer» As, `lambda=h/(sqrt(2mqV)) :. lambda prop 1/(sqrt(mq)) :. (lambda_(p))/(lambda_(alpha)) =(sqrt(m_(alpha)q_(alpha)))/(sqrt(m_(p)q_(p)))=(sqrt(4m_(p)xx2e))/(sqrt(m_(p)xxe))=sqrt8=2sqrt2` | |
| 366. |
Find the de-Broglie wavelength (in Å) assciated with a photon moving with a velocity 0.5 c, where `c=3xx10^(8)m//s`, rest mass of proton `=1.675xx10^(-27)kg, h=6.6xx10^(-34)Js`. |
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Answer» Correct Answer - `2.27xx10^(-15) m` Rest mass proton `m_(0)=1.675xx10^(-27)kg` `v=0.5 c=0.5xx(3xx10^(8))=1.5xx10^(8)m//s` As v is comparable to c, hence mass of proton in motion will be a relative mass. So, `m=(m_(0))/(sqrt(1-v^(2)//c^(2)))=(m_(0))/(sqrt(1-(0.5c)^(2)//c^(2))` `=(m_(0))/(sqrt(0.75))=(2m_(0))/(sqrt(3))` `lambda=h/(mv)=(6.63xx10^(-34))/((2m_(0)//sqrt(3))xxv)` `=(6.6xx10^(-34)xxsqrt(3))/(2xx1.657xx10^(-27)xx(1.5xx10^(8)))` `=2.27xx10^(-5)m` |
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| 367. |
An `alpha`-particle and a proton are accelerated from rest through the same potential difference V. Find the ratio of de-Broglie wavelength associated with them. |
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Answer» Correct Answer - `1//(2sqrt2)` The kinetic energy gained by a charged particle of charge q, mass m, moving with velocity v, when acceleration through the potential difference V is K.E. `=1/2mv^(2)=qV or mv=[2mqV]^(1//2)` De-Broglie wavelength of the particle is `lambda=h/(mv)=h/([2mqV]^(1//2))` so `lambda prop 1/(sqrt(mq))` (for the same value of V) `:. (lambda_(alpha))/(lambda_(p))=sqrt((m_(p)q_(p))/(m_(alpha)q_(alpha)))=sqrt((m_(p)xxe)/(4m_(p)xx2e)) =1/(sqrt(8)) =1/(2sqrt(2))` |
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| 368. |
A particle with rest mass `m_0` is moving with velocity c. what is the de-Broglie wavelength associated with it? |
| Answer» `lambda=h/(mv)=(h sqrt(1-v^2//c^2))/(m_0v)=(h sqrt(1-c^2//c^2))/(m_0c)=0` | |
| 369. |
Find the de-Broglie wavelength (inÅ) associated with an electron moving with a velocity 0.6c, where `c=3xx10^8 m//s` and rest mass of electron `=9.1xx10^(-31)kg, h=6.63xx10^(-34)Js`. |
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Answer» Rest mass of electron, `m_0=9.1xx10^(-31)kg` `v=0.6c=0.6xx3xx10^8=1.8xx10^8ms^-1` As v is comparable to c, hence mass of the electron in motion will be a relativistic mass. So, `m=(m_0)/(sqrt(1-v^2//c^2))=(m_0)/(sqrt(1-((0.6c)^2)/(c^2)))=(m_0)/0.8` De-broglie wavelength, `lambda=h/(mv)=h/((m_0//0.8)xxv)=(hxx0.8)/(m_0v)` `=((6.6xx10^(-34))xx0.8)/((9.1xx10^(-31))xx(1.8xx10^8))` `=0.322xx10^(-11)m=0.0322xx10^(-10)m` `0.032Å` |
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| 370. |
The work function of a surface of a photosensitive material is `6.2 eV`. The wavelength of the incident radiation for which the stopping potential is `5 V` lies in theA. ultraviolet regionB. visible regionC. infrared regionD. X - ray region |
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Answer» Correct Answer - A According to laws of photoelectric effect `KE_(max) = E - phi` where `phi` is work function and `KE_(max)` , is maximum , kinetic energy of photoelectron. `:. Hv = eV_(0) + phi` or `hv = 5 eV + 6.2 eV = 11.2 eV` `:. Lambda = ((12400)/(11.2)) Å ~~ 1000 Å` Hence , the radiation lies in ultraviolet region. |
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| 371. |
Find the minimum wavelngth of `X` - ray emitted by `X` - ray tube , which is operating at `15 kV` accelerating voltage.A. `0.75 Å`B. `0.82 Å`C. `1.42 Å`D. `1.13 Å` |
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Answer» Correct Answer - B `lambda_(min) = (12400)/(V_((i n v o l t))) = (12400)/( 15 xx 10^(3)) = 0.82 Å` |
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| 372. |
From your answer to (a) , guess what order of accelerating voltage ( for electrons ) is required in such a tube ? |
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Answer» In X - ray tube , accelerating voltage provides the energy to the electrons which produce X - rays . For getting X - rays , photons of of 27.51 KeV is required that the incident electrons must posess kinetic energy of 27.61 KeV Energy = eV = E ,eV= 27.6 KeV, V = 27.6 KV So , the order of accelerating voltage is 30 KV |
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| 373. |
Assuming photoemission to take place , the factor by which the maximum velocity of the emitted photoelectrons changes when the wavelength of the incident radiation is increased four times , isA. `4`B. `(1)/(4)`C. `2`D. `(1)/(2)` |
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Answer» Correct Answer - D `(hc )/(lambda) = W_(0) + (1)/(2) mv_(max)^(2)`. Assuming `W_(0)` to be negligible in comparison to `(hc )/(lambda)` i.e., `v_(max)^(2) prop (1)/(lambda) rArr v_(max) prop (1)/(sqrt(lambda))`. (On increasing wavelength `lambda to 4 lambda , v_(max)` becomes half). |
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| 374. |
In an accelerator experiment on high energy collision of electrons with positrons, a certain events is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two `gamma`-rays of equal energy. What is the wavelength associated with each `gamma`-rays? `(1BeV=10^(9)eV)` |
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Answer» Given, energy of 2 `gamma`- rays 2E = 10.2 BeV `[because overset(0)underset(-1)e+overset(0)underset(+1)e rarr2gamma -" rays "]` `rArr 2(hc)/(lambda)=10.2BeV[becauseE=(hc)/(lambda)]rArr lambda=(2hc)/(10.2BeV)` Here h = ` 6.63 xx 10^(-34) J - S, c = 3 xx 10^(8) ` m/s , 1 BeV = ` 10^(9) eV = 10^(9 ) xx 1.6 xx 10^(-19) ` J ` rArr lambda = ( 2 xx 6.63 xx 10^(-34) xx 3xx 10^(8))/(10.2 xx 10^(9) xx 1.6 xx 10^(-19)):. lambda = 2.436 xx 10^(-16) ` m |
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| 375. |
Consider fig. for photoemission. How would you recocile with momentum-conservation? Note light (photons) have momentum in a different direction than the emitted electrons. |
| Answer» During photoelectric emission, the momentum of incident photon is transfered to the metal. At microscopic level, atoms of a metal absorb the photon and its momentum is transfered mainly to the nucleus and electrons. The electron excited is emitted. Therefore, the conservation of momentum is to be considerd as the momentum of incident photon transfered to the nucleus and electrons. | |
| 376. |
In a photon-particle collision (sauch as photon-electron collision), which of the following may not be conserved ?A. Total energyB. Number of photonsC. Total momentumD. both (a) and (b) |
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Answer» Correct Answer - B (b) : In a photon-particle collision, (such as photon electron collision), the total energy and total momentum are conserved. However, the number of photons may not be conserved in photon-particle collision. The photon may be absorbed or a new photon may be created. |
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| 377. |
Assuming an electron is confined to a 1nm wide region, find the wavelength in momentum using Heisenberg Uncetainity principal `(Deltax Deltap~~h)`. You can assume the uncertainity in position `Deltax` and 1nm. Assuming `p~=Deltap`, find the energy of the electron in electron volts. |
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Answer» Here, `Deltax=1nm=10^(-9)m, Deltap= ?` As, `DeltaxDeltap~~h :. Deltap=h/(Deltax)=h/(2piDeltax) =(6.62xx10^(-34)Js)/(2xx(22//7)10^(-9)m)=1.05xx10^(-25)kgms^(-1)` Energy, `E=(p^(2))/(2m)=((Deltap)^(2))/(2m)=((1.05xx10^(-25))^(2))/(2xx9.1xx10^(-31))J = ((1.05xx10^(-25))^(2))/(2xx9.1xx10^(-31)xx1.6xx10^(-19)) eV=3.8xx10^(-2)eV` (Given `p= Deltap`) |
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| 378. |
The rest mass of the photon isA. `(hv)/(c )`B. `(hv)/(c^(2))`C. `(hv)/(lamda)`D. zero |
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Answer» Correct Answer - D (d) : The rest mass of photon is zero. |
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| 379. |
On what principal is an electron micro-scope based? |
| Answer» An electron microscope is based on de-Broglie hypothesis. According to it, a beam of electrons behaves as a wave which can be converged or diverged by magnetic or electric field lenses like a beam of light using optical lenses. | |
| 380. |
The minimum energy required to ejected an electron from the surface of a metal is called............ . |
| Answer» Correct Answer - threshold frequency | |
| 381. |
Ultraviolet light of wavelength 66.26 nm and intensity `2 W//m^(2)` falls on potassium surface by which photoelectrons are ejected out. If only 0.1% of the incident photons produce photoelectrons, and surface area of metal surface is `4 m^(2)`, how many electrons are emitted per second?A. `2.67xx10^15`B. `3xx10^15`C. `3.33xx10^17`D. `4.17xx10^16` |
| Answer» Correct Answer - 1 | |
| 382. |
The linear momentum of a 3 MeV photon isA. `0.01eVsm^(-1)`B. `0.02eVsm^(-1)`C. `0.03eVsm^(-1)`D. `0.04eVsm^(-1)` |
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Answer» Correct Answer - A (a) : Energy of a photon, `E=3MeV=3xx10^(6)eV` Linear momentum of the photon, `p=(E)/( c)` where c is the speed of light in vacuum `p=(3xx10^(6)eV)/(3xx10^(8)"m s"^(-1))=10^(-2)eV"s m"^(-1)=0.01eV"s m^(-1)` |
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| 383. |
Number of ejected photoelectorns increases with an increase in intensity of light but not with the increase in frequency of light. Why? |
| Answer» One incident photon can eject one photoelectron from a photosestive surface. Therefore the no. of photoelectrons ejected per second depends upon the no. of photons falling per second on it, which in turns depends upon the intensity of the incident light. The increase in the frequency of the incident photon but one photon of high energy can not ejected more than one photoelectron from a photosenstive surface. | |
| 384. |
Calculate the strength of the transverse magnetic field required to bend all photoelectorns within a circle of radius 50 cm, when light of wavelength 4000Å is incident on a barium emitter. Work function of barium is 2.5eV. Given, `m_e=9.1xx10^(-31)kg, e=1.6xx10^(-19)C, h=6.6xx10^(-34)Js.` |
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Answer» Here, B=? , r=50cm=0.50m, `lambda=4000Å=4000xx10^(-10)m=4xx10^-7m,` `phi_02.5eV=2.5xx1.6xx10^(-19)J`. As, `1/2mv_(max)^2=(hc)/lambda-phi_0` or `v_(max)^2=2/m[(hc)/lambda-phi_0]` `=2/(9.1xx10^(-31))` `xx[((6.6xx10^(-34))xx(3xx10^8))/(4xx10^-7)-2.5xx1.6xx10^(-19)]` `=20.88xx10^(10)` `:. v_(max)=4.75xx10^5m//s` Now, `Bev_(max)=mv_(max)^2//r` or `B=(mv_(max))/(er)=((9.1xx10^(-31))xx(4.57xx10^5))/((1.6xx10^(-19))xx0.50)` `5.2xx10^-6T` |
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| 385. |
Light of wavelength 4000Å is incident on barium. Photoelectrons emitted describe a circle of radius 50 cm by a magnetic field of flux density `5.26xx10^-6 tesla`. What is the work function of barium in eV? Given `h=6.6xx10^(-34)Js , e=1.6xx10^(-19)C, m_e=9.1xx10^(-31)kg.` |
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Answer» `v_"max"^2=2/m["hc"/lambda-phi_0]` `=2/(9.1xx10^(-31))[((6.6xx10^(-34))xx(3xx10^8))/(4xx10^(-7))-2.5xx1.6xx10^(-19)]` `=20.88xx10^10` `therefore v_"max"=4.57xx10^5` m/s `Bev_"max"=(mv_"max"^2)/r` `rArr r=(mv_"max")/"Be"` `=((9.1xx10^(-31))xx(4.57xx10^5))/(5.2xx10^(-6)xx1.6xx10^(-19))=1/2 m` |
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| 386. |
Light of wavelength 4000Å is incident on barium. Photoelectrons emitted describe a circle of radius 50 cm by a magnetic field of flux density `5.26xx10^-6 tesla`. What is the work of barium in eV? Given `h=6.6xx10^(-34)Js , e=1.6xx10^(-19)C, m_e=9.1xx10^(-31)kg.` |
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Answer» Here, `lambda=4000xx10^(-10)m=4xx10^-7`, `r=50cm =0.50m, B=5.26xx10^-6T`. Force on electron in magnetic field is, `F=Bev=(mv^2)/r or v=(Ber)/m` Work function of barium is given by `phi_0=(hc)/lambda-1/2mv^2=(hc)/lambda-1/2mxx((Ber)/(m))^2` `=(hc)/lambda-1/2 (B^2e^2r^2)/m` `=((6.6xx10^(-34))xx(3xx10^8))/((4xx10^-7)xx(1.6xx10^(-19))` `-1/2 xx((5.26xx10^-6)^2xx(1.6xx10^(-19))^2xx(0.50)^2)/((9.1xx10^(-31))xx(1.6xx10^(-19)))` `~~2.5 eV` |
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| 387. |
Two metallic plates P (collector) and Q (emitter) are separated by a distance of 0.10 meter. These are connected through an ammeter without any cell. A magnetic field B exists parallel to the plates. Light of wavelength between `4000Å and 6000Å` fall on the plate whose work function is 2.39eV. Calculate the minimum value of B for which the current registered with ammeter is zero. Use, `h=6.6xx10^(-34)Js, e=1.6xx10^(-19)C` |
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Answer» Energy of the incident photon of light of wave length `6000Å` is `E=(hc)/lambda=((6.6xx10^(-34))xx(3xx10^(8)))/(6000xx1.6xx10^(-19))eV=2.01eV` Since this value of energy is less than work function `(=2.39eV)` of the metal, hence no photoelectron is emitted for light of wavelength `6000Å` Since the energy of photon of light of wavelength `4000Å` is greater than 2.39 eV , so photoelectric emission takes place with its helt. Maximum K.E. of the emitted photoelectron is `K_(max)=1/2mv_(max)^(2)=(hc)/lambda-phi_(0)= ((6.6xx10^(-34))xx(3xx10^(8)))/(4000xx10^(-10)xx1.6xx10^(-19))-2.39eV` `=3.1-2.39=0.71 eV=0.71xx1.6xx10^(-19)J` `v_(max)=[(2xx0.71xx1.6xx10^(-19))/(9.1xx10^(-31))]^(1//2)=5xx10^(5)m//s` For zero constantl, `(mv^(2))/r=evB or B=(mv)/(er)=((9.1xx10^(-31))xx5xx10^(5))/((1.6xx10^(-19))xx0.1)=2.86xx10^(-5)T` |
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| 388. |
Two metallic plates `A and B` , each of area `5 xx 10m` are placed parallel to each other at a separation of 1 cm . Plate B carries a positive charge of 33.7 pc . A monochromatic beam of light, with photons of energy 5 eV each, starts falling on plate A at t = 0, so that 10 photons fall on it per square meter per second. Assume that one photoelectron is emitted for every 10 incident photons. Also assume that all the emitted photoelectrons are collected by plate B and the work function of plate A remains constant at the value 2 eV . Electric field between the plates at the end of 10 seconds isA. `5xx10^(6)`B. `7xx10^(7)`C. `5xx10^(7)`D. `9xx10^(6)` |
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Answer» Correct Answer - C (c ) : Area of each plate `=xx10^(-14)m^(2)` Separation between plates `=cm=10^(-2)m` `("Photons incident")/("area"xx"time")=10^(16),t=10s` `:."No. of photons incident"=10^(16)xx"area"xx"time"=10xx5xx10^(13)` `:."Required number of photoelectrons emitted"=(5xx10^(13))/(10^(6))` Number of photoelectrons `=5xx10^(7)` . . .(i) |
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| 389. |
Monochromatic light of wavelength `3000 Å` is incident on a surface area `4 cm^(2)`. If intensity of light is `150 m W//m^(2)` , then rate at which photons strike the target is |
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Answer» `l=(nhc)/(lambdaA)xx1/t` (Intensity is energy crossing per unit area per unit time ) `n/t=(l lambdaA)/(hc)` `=(150xx10^(-3)xx4xx10^(-4)xx3xx10^(-7))/(6.6xx10^(-34)xx3xx10^8)` `=9xx10^13` per second |
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| 390. |
Two large parallel plates are connected with the terminal of `100 V` power supply. These plates have a fine hole at the centre . An electron having energy `200 eV` is so directed that it passes through the holes . When it comes out its de - Broglie wavelength is A. `1.22 Å`B. `1.75 Å`C. `2 Å`D. None of these |
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Answer» Correct Answer - A Energy of the electron , when it comes out from the second plate ` = 200 eV - 100 eV = 100 eV` Hence accelerating potential difference ` = 100 V` `lambda_(el ectron) = (12.27)/(sqrt(V)) = (12.27)/(sqrt(100)) = 1.23 Å` |
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| 391. |
The wavelength associated with an electron accelerated through a potential difference of `100 V` is nearly |
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Answer» if the kinetic energy is much less than the rest energy , we can use classical `KE=1/2mv^2` For an electron, `m_0 c^2`=0.511 MeV. We then apply conservation of energy : the KE acquired by the electron equals its loss in PE. After solving for v, we use `lambda=h/"mv" ` to that the de Broglie wavelength . The gain in kinetic energy will equal the loss in potential energy (`DeltaPE`=eV-0) : KE=eV. so KE=100 eV. The ratio `"KE"/(m_0c^2)="100 eV"/((0.511xx10^6 eV))approx 10^(-4)` , so relativity is not needed. Thus `1/2mv^2=eV` and `v=sqrt((2 eV)/m)` `sqrt(((2)(1.6xx10^(-19)C)(100 V))/((9.1xx10^(-31) kg)))` `=5.9xx10^6` m/s Then `lambda=h/"mv"` `=((6.63xx10^(-34) J.s))/((9.1xx10^(-31) kg)(5.9xx10^6 m//s))` `=1.2xx10^(-10)` m or 0.12 nm |
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| 392. |
Which of the following figure represents the variation of particle momentum and the associated de - Broglie wavelength ?A. B. C. D. |
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Answer» Correct Answer - D de - Broglie wavelength `lambda = (h)/(p) rArr lambda prop (1)/(p)` i.e., graph will be a rectangular hyperbola. |
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| 393. |
The wavelength associated with an electron accelerated through a potential difference of `100 V` is nearlyA. `100 Å`B. `123 Å`C. `1.23 Å`D. `0.123 Å` |
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Answer» Correct Answer - C `lambda = (1)/(sqrt( 2 m Q V))` ` = (6.6 xx 10^(-34))/(sqrt(2 xx 9.1 xx 10^(-31) xx 1.6 xx 10^(-19) xx 100))` `= 1.23 Å` |
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| 394. |
Davisson and Germer experiment provedA. Wave nature of lightB. Particle nature of lightC. Both (a) and (b)D. Neither (a) nor (b) |
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Answer» Correct Answer - D Davisson and Germer proved the wave nature of electron by performing an experiment. |
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