## DUAL NATURE OF RADIATION AND MATTER quiz-12

As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .

Q1. Photons of energy of 6 eV are incident on a metal surface whose work function is 4 eV. The minimum kinetic energy of the emitted photoelectrons will be
•  Zero
•  1 eV
•  2 eV
•  10 eV
Solution
1/2 mv2=hv-Ï•0=hv-hv0 For minimum kinetic energy of emitted photoelectron, v=v0 ∴ 1/2 mv2=0

Q2.A ratio transmitter operates at a frequency 880 kHz and a power of 10 kW. The number of photons emitted per second is
•  1.72×1031
•  1.327×1025
•  1.327×1037
•  1.327×1045
Solution
Number of photons emitted per sec, n=Power/(Energy of photon)=P/hv =10000/(6.6×10-34×880×103 )=1.72×1031

Q3. An important spectral emission line has a wavelength of 21cm. The corresponding photon energy is (h=6.62×10-34 Js and c=3×108 ms-1)
•   5.9×10-8 eV
•  5.9×10-4 eV
•  5.9×10-6 eV
•  11.8×10-6 eV
Solution
E=hv/Î» = hc/eÎ» (ineV) =(6.6×10-34×3×108)/(1.6×10-19×0.21)=5.9×10-6 eV

Q4. If 10000 V is applied across an X-ray tube, what will be the ratio of de-Broglie wavelength of the incident electrons to the shortest wavelength of X-ray produced (e/m for electron is 1.8×1011 ckg-1)
•  1
•  0.1
•  0.2
•  0.3
Solution
For incident electron 1/2 mv2=eV orp2=2meV ∴de-Broglie wavelength Î»1=h/p=h/√2meV Shortest X-ray wavelength Î»2=hc/eV ∴Î»1/Î»2 =1/c √((V/2)(e/m) )=√(104/2×1.8×1011 )/(3×108 )=0.1

Q5.In an experiment of photoelectric effect the stopping potential was measured to be V1 and V2 volts with incident light of wavelength Î» and Î»/2 respectively. The relation between V1 and V2 may be
•  V2<V1
•  V1<V2<2V1
•  V2=2V1
•  V2>2V1
Solution
According to Einstein’s photoelectric equation Kmax=hv-Ï•0 eVs=hc/Î»-Ï•0⇒Vs=hc/Î»e-Ï•0/e Where, Î»= Wavelength of incident light Ï•0= Work function Vs= Stopping potential According to given problem V1=hc/Î»e-Ï•0/e V2=hc/(Î»/2)e-Ï•0/e V2=2hc/Î»e-Ï•0/e=2hc/Î»e-(2Ï•0)/e+(2Ï•0)/e-Ï•0/e =2(hc/Î»e-Ï•0/e)+Ï•0/e V2=2V1+Ï•0/e [Using (i)] ∴V2>2V1

Q6. A light of wavelength 4000 â„« is allowed to fall on a metal surface having work function 2 eV. The maximum velocity of the emitted electrons is (R=6.6×10-34Js)
•  1.35×105 ms-1
•  2.7×105 ms-1
• 6.2×105 ms-1
•  8.1×105 ms-1
Solution
1/2 mv2=hc/Î»-Ï• (in eV) =(6.6×10-34×3×108)/(4000×10-10×1.6×10-19 )-2 =3.1-2=1.1 eV=1.1×1.6×10-19 J =1.76×10-19 J v=(1.76×10-19×2)/(9×10-13 ) =6.2×105 ms-1

Q7.The stopping potential (V0) versus frequency (v) plot of a substance is shown in figure the threshold wave length is

•  5×1014 m
•  6000 â„«
•  5000 â„«
•  Cannot be estimated from given data
Solution
Î»0=c/v0 =(3×108)/(5×1014 )=6×10-7 m=6000 â„«

Q8.The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is (Radius of the first orbit of hydrogen atom =0.53â„« )
•  1.67 â„«
•  3.33 â„«
•  1.06 â„«
•  0.53 â„«
Solution
According to Bohr’s quantisation of angular momentum mvr=nh/2Ï€ Or h/mv=2Ï€r/n ..(i) de-Broglie wavelength Î»=h/mv ..(ii) From Eqs. (i) and (ii), we get Wavelength Î»=2Ï€r/n =(2×Ï€×0.53 â„«)/1=3.33â„«

Q9.The ratio of the energy of a photon with Î»=150 nm to that with Î»=300 nm is
•  2
•  1/4
•  4
•  1/2
Solution
According to Einstein, the energy of photon is given by E=hv=hc/Î» Where h is Planck’s constant, c the speed of light and Î» the wavelength. ∴ E1/E2 =Î»2/Î»1 Given, Î»1=150 nm ,Î»=300nm ∴ E1/E2 =300/150=2/1

Q10. X-rays when incident on a metal
•  Exert a force on it
•  Transfer energy to it
•  Transfer pressure to it
• All of the above
Solution
All of the above

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