InterviewSolution
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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 momentum of a photon having energy equal to the rest energy of an electron isA. zeroB. `2.73 xx 10^(-22)kg ms^(-1)`C. `1.99 xx 10^(-24)kg ms^(-1)`D. infinite |
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Answer» Correct Answer - B For photon, `p=(h)/(lambda)=(E)/(c )` But from questions, E = rest energy of electron = `m_(0)c^(2)` `thereforep=(m_(0)c^(2))/(c )=m_(0)c=(9.109xx10^(-31)kg)xx(3xx10^(8)ms^(-1))` `=2.7327 xx 10^(-22)kgms^(-1)=2.73xx10^(-22)kg m//s` |
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| 2. |
In an oil experiment, the following charges (in arbitrary units) were found on a series of oil droplets:A. `2.30xx10^(-15)`B. `1.15xx10^(-15)`C. `1.38xx10^(-14)`D. `1.955xx10^(-14)` |
| Answer» Correct Answer - B | |
| 3. |
A moving electron has numerical relation `lambda=h`. Then,A. `m_(e)=(1)/(v_(e))`B. `v_(e)=(1)/(m_(e))`C. both a and bD. None of these |
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Answer» Correct Answer - C `lambda=(h)/(m_(e)v_(e))` `therefore " "m_(e)v_(e)=1 " "[becauselambda=h("given")]` `therefore" "m_(e)=(1)/(v_(e))` |
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| 4. |
An `alpha`-particle when accelerated through a potential of V volt has a wavelength `lambda` associated with it, but if a proton in order to have same wavelength `lambda` by what potential difference it must be accelerated?A. 8 VB. 6 VC. 4 VD. 12 V |
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Answer» Correct Answer - A `lambda_(p)=lambda_(alpha)` `(h)/(sqrt(2m_(p)Q_(p)V_(p)))=(h)/(sqrt(2m_(alpha)Q_(alpha)V_(alpha)))` `therefore" "m_(p)Q_(p)V_(p)=m_(alpha)Q_(alpha)V_(alpha)` `therefore" "V_(p)=((m_(alpha))/(m_(p)))((Q_(alpha))/(Q_(p)))V_(alpha` `V_(p)=(4)(2)V_(alpha)implies V_(p)=8V` |
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| 5. |
The voltage applied to an electron microscope to produce electrons of wavelength `0.50 Å` isA. 602 VB. 50 VC. 138 VD. 812 V |
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Answer» Correct Answer - A de-Broglie wavelength is `lambda=(h)/(mv)=(h)/(sqrt(2mE))` `"But "E=ev` `lambda=(h)/(sqrt(2meV))implies V=(h^(2))/(2melambda^(2))` `V=((6.62 xx 10^(-34))^(2))/((0.5xx10^(-10))^(2)xx2xx9.1xx10^(-31)xx1.6xx10^(-19))` `impliesV=601.98V=602V` |
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| 6. |
If the velocity of an electron is doubled, its de-Broglie frequency will beA. halfB. remain sameC. doubledD. become four times |
| Answer» Correct Answer - C | |
| 7. |
The cathode ray particles originate in a discharge tube from theA. cathodeB. anodeC. source of high voltageD. residual gas |
| Answer» Correct Answer - A | |
| 8. |
A surface is irradiated with ultraviolet radiation of wavelength `0.2 mum`. If the maximum velocity of electron liberated from the surface is `8.8 xx 10^(5)m//s`, then find the work function of the surface.A. 3 eVB. 4 eVC. 5 eVD. 6 eV |
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Answer» Correct Answer - D We know that E = hv = `(hc)/(lambda)` `E=(6.6xx10^(-34)xx3xx10^(8))/(0.2 xx10^(-6))` `=(1.98xx10^(-25))/(0.2 xx10^(-6))=9.9xx10^(-19)J` Also, `" "E-W_(0)=(1)/(2)mv^(2)` `9.9xx10^(-19)-W_(0)=(1)/(2)xx9.1xx10^(-31)(8.8xx10^(5))^(2)` `W_(0)=9.9xx10^(-19)-3.52xx10^(-19)` `W_(0)=6.38~~ 6 eV` |
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| 9. |
An X-ray tube is operating at 15 kV. The lower limit of the wavelength of X-ray produced isA. `0.82 xx 10^(-7)m`B. `0.82 xx 10^(-8)m`C. `0.83 xx 10^(-10)m`D. `0.82 xx 10^(-13) m` |
| Answer» Correct Answer - C | |
| 10. |
If `n_(r )` and `n_(b)` are the number of photons of red and blue light respectively with same energy, thenA. `n_(r ) gt n_(b)`B. `n_(r ) lt n_(b)`C. `n_(r ) = n_(b)`D. no relation between `n_(r )` and `n_(b)` |
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Answer» Correct Answer - A `("VIBGYOR")/(lambda"(increases)")E=n.(hc)/(lambda)` From equation, `" "E_(r )=E_(b)" (Given)"` `" or "n_(r ).(hc)/(lambda_(r ))=n_(b).(hc)/(lambda_(b))" or "(n_(r ))/(lambda_(b))=(lambda_(r))/(lambda_(b))because(lambda_(r ))/(lambda_(b))gt1` Hence, `(n_(r ))/(n_(b))gt1" or "n_(r )gt n_(b)` |
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| 11. |
Photons of an electromagnetic radiation has an energy 11 keV each. To which region of electromagnetic spectrum does it belong ?A. X-ray regionB. Ultra violet regionC. Infrared regionD. Visible region |
| Answer» Correct Answer - B | |
| 12. |
At its closet approach, the distance between the mars and the earth is found to be 60 million km. When the planets are at this closet distance, how long would it take to send a radio message from a space probe of mars to earth?A. 5 sB. 200 sC. 0.2 sD. 500 s |
| Answer» Correct Answer - B | |
| 13. |
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. `2.67 xx 10^(15)`B. `3 xx 10^(15)`C. `3.33 xx 10^(7)`D. `4.17 xx 10^(16)` |
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Answer» Correct Answer - A Number of photons falling on metal surface, `n_(p)=("intensity"xx"area")/("energy per quanta")` `=((2Js^(-1)m)xx(4m^(2)))/({{(6.626xx10^(-34)Js)xx(3xx10^(8)ms^(-1)))/((66.26xx10^(-9)m)}})` `=2.67xx10^(18)" per sec"` From equation, `n_(e)=0.1%" of "n_(p)` `=(0.1)/(100)xx2.67xx10^(18)=2.67xx10^(15)" per sec"` |
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| 14. |
In previous problem, the stopping potential isA. 1.9 VB. 10 VC. 3 VD. None |
| Answer» Correct Answer - A | |
| 15. |
`10^-3` W of `5000A` light is directed on a photoelectric cell. If the current in the cell is `0.16muA`, the percentage of incident photons which produce photoelectrons, isA. 0.004B. 0.0004C. 0.2D. 0.1 |
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Answer» Correct Answer - B Current, `l=(q)/(t)=(Ne)/(t)=n_(e)e` where, `n_(e)` = number of electrons per second `e = 1.6 xx 10^(19)C` As, each electron is emitted due to absorption of a photon, `therefore n_(e)=n_(2)` = number of photons absorbed per second `implies " "n_(r)=(l)/(e)=(0.16xx10^(-6))/(1.6xx10^(-19))"per sec"= 10^(12)" photon"//s` And power of source, `P=n_(1)E, n_(1)` = number of photons emitted per second `implies" "n_(1)=(P)/(E)` `"where, "E=(hc)/(lambda)=(12400)/(5000)eV` `impliesE=2.48eV=2.48xx1.6 xx10^(-19)J=3.968 xx10^(-19)J` `impliesn_(1)=(P)/(E)=2.52xx10^(15)" photons"//s` `therefore %` of photons absorbed = `(n_(2))/(n_(1))xx100` `=(10^(12)xx100)/(2.52xx10^(15))=0.03968% ~~0.04%` |
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| 16. |
A red bulb and violet bulb of equal power emits `n_(R )` and `n_(v)` number of photons in a given time, thenA. `n_(R ) = n_(v)`B. `n_(R ) gt n_(v)`C. `n_(R ) lt n_(v)`D. `n_(R ) ge n_(v)` |
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Answer» Correct Answer - B We know that, `E-(hc)/(lambda)` Let, power=P then, energy in time t `nE=Pt" and "n=(Pt)/(E)=(Ptlambda)/(hc)` `therefore" "n_(R)=(Ptlambda_(R))/(hc)" "...(i)` `n_(v)=(Ptlambda_(v))/(hc)" "....(ii)` Dividing Eq. (i) by Eq. (ii), we get `(n_(R))/(n_(v))=(lambda_(R))/(lambda_(v))` We know, `lambda_(R) gt lambda_(v)` `n_(R) gt n_(V)` |
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| 17. |
Only a fraction of the electrical energy supplied to a tungsten light bulb is converted into visible light. If a 100 W light bulb converts 20% of the electrical energy into visible light `(lambda=662.6nm)`, then the number of photons emitted by the bulb per second isA. `6.67 xx 10^(19)`B. `2xx10^(28)`C. `6xx10^(36)`D. `30xx10^(19)` |
| Answer» Correct Answer - A | |
| 18. |
A radiation is incident on a metal surface of work function 2.3 eV. The wavelength of incident radiation is 60 nm, then the number of photoelectrons isA. zeroB. `gt 10^(4)`C. `=10^(4)`D. None of these |
| Answer» Correct Answer - A | |
| 19. |
A and B are two metals with threshold frequencies `1.8 xx 10^(14)Hz` and `2.2 xx 10^(14)Hz`. Two identical photons of energy 0.825 eV each are incident on them. Then, photoelectrons are emitted by (Taking, `h = 6.6 xx 10^(-34)J-s`) |
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Answer» Correct Answer - B Threshold energy of A is `E_(A)=hv_(A)` `=6.6xx10^(-34)xx1.8 xx10^(14)` `11.88xx10^(-20)J` `=(11.88xx10^(-20))/(1.6 xx 10^(-19))eV` `=0.74 eV` Similarl, `E_(B) = 0.91 eV` Since, the incident photons have energy greater than `E_(A)` but less than `E_(B)`. So, photoelectrons will be emitted from metal A only. |
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| 20. |
Threshold frequency for photoelectric effect from a metal surface of work function 4.5 eV isA. `1.1 xx 10^(9) Hz`B. 540 HzC. `1.1 xx 10^(15)` HzD. None of these |
| Answer» Correct Answer - C | |
| 21. |
The work function of a certain metal is 2.3 eV. If light of wave number `2 xx 10^(6)m^(-1)` falls on it, the kinetic energies of fastest and slowest ejected electron will be respectively.A. 2.48 eV, 0.18 eVB. 0.18 eV, zeroC. 2.30 eV, 0.18 eVD. 0.18 eV, 0.18 eV |
| Answer» Correct Answer - B | |
| 22. |
Calculate the number of photons emitted by a 60 W bulb per second, if 10% of the electrical energy supplied to an incandescent light bulb is radiated as visible light.A. `1.8 xx 10^(19)`B. `1.8 xx 10^(16)`C. `1.8 xx 10^(11)`D. `1.8 xx 10^(21)` |
| Answer» Correct Answer - A | |
| 23. |
A certain molecule has an energy level diagram for its vibrational energy in which two levels are 0.014 eV apart. Find the wavelength of the emitted line for the molecule as it falls form one of these levels to the otherA. `8.9xx10^(-5)m`B. `1.2xx10^(-6)m`C. 173.6 mD. `4.6 xx 10^(-7)m` |
| Answer» Correct Answer - A | |
| 24. |
At one time, the metre was defined as 1650763.73 wavelengths of the orange light emitted by a light source containing `Kr^(86)` atoms. What is the corresponding photon energy of this radiation?A. `3.28 xx 10^(-19)` J/quantaB. `1.204 xx 10^(-31)` J/quantaC. `1.09xx10^(-27)` J/quantaD. `4.01 xx 10^(-40)` J/quanta |
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Answer» Correct Answer - A Given, `bar(v)=1650763.73m^(-1)` Now, `E=hcbar(v)` `=(6.626xx10^(-34))xx(3xx10^(8))xx1650763.73` `=3.28 xx10^(-19)J//"quanta"` |
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