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.**A beam of 35.0 keV electrons strikes a molybdenum target, generating the X-rays. What is the cut-off wavelength?

Solution

The cut-off wavelength Î»

The cut-off wavelength Î»

_{min}corresponds to an electron transferring (approximately) all of its energy to an X-ray photon, thus producing a photon with the greatest possible frequency and least possible wavelengh. From relation Î»_{min}=hc/K_{o}=(4.14×10^{-15})(3×10^{8})/(35.0×10^{3}) = 3.55×10^{-11}m=35.5pm**Q2.**The wavelength of a 1 keV photon is 1.24 nm. The frequency of 1 MeV photon is

Solution

f=c/Î»=c/(hc/E)=E/h ∴ f=(1×1.6×10

f=c/Î»=c/(hc/E)=E/h ∴ f=(1×1.6×10

^{-13})/(6.6×10^{-34})=2.4×10^{20}Hz**Q3.**The minimum wavelength of X-ray emitted from X-ray machine operating at an accelerating potential of V volts is

Solution

If all of the kinetic energy carried by an electron is converted into radiation, the energy of the X-rays photon would be given by E

If all of the kinetic energy carried by an electron is converted into radiation, the energy of the X-rays photon would be given by E

_{max}=hv_{max}=eV Where h is Planck’s constant, v_{max}the largest frequency, e charge of an electron and V the applied voltage. This maximum energy or minimum wavelength is called the Duane-Hunt limit. ∴ hv_{max}=hc/Î»_{min}=eV ⇒ Î»_{min}=hc/eV

**Q4.**If the energy of a photon corresponding to a wavelength of 6000 â„« is 3.32×10

^{-19}J, the photon energy for a wavelength of 4000 â„« will be

Solution

E=hc/Î»⇒E

E=hc/Î»⇒E

_{1}/E_{2}=Î»_{2}/Î»_{1}⇒(3.32×10^{-19})/E_{2}=4000/6000 ⇒E_{2}=4.98×10^{-19}J=3.1 eV**Q5.**An X-ray pulse of wavelength 4.9 â„« is sent through a section of Wilson cloud chamber containing a super saturated gas, and tracks of photoelectron ejected from the gaseous atoms are observed. Two groups of tracks of lengths 1.40 cm and 2.02 cm are noted. If the range-energy relation for cloud chamber is given by R=Î±E with Î±=1cm/keV, obtain the binding energies of the two levels from which electrons are emitted. Given h=6.63×10

^{-34}J-s,e=1.6×10

^{-19}J

Solution

Binding energy, W=hc/Î»-E; Where E=R/Î±; Here,Î»=4.9×10

Binding energy, W=hc/Î»-E; Where E=R/Î±; Here,Î»=4.9×10

^{-10}m R_{1}=1.4cm,R_{2}=2.02cm; Î±=1cm/keV For cloud chamber, the range-energy relation is R=Î±E orE=R/Î± ∴E_{1}=R_{1}/Î±=1.40/1=1.40keV and E_{2}=R_{2}/Î±=2.02/1=2.02keV Energy of the incident photon hv=hc/Î»=(6.63×10^{-34}×3×10^{8})/(4.9×10^{-10}) J=2.54 keV From Einstein’s Photoelectric equation hv=W+E ∴ Binding energy,W=hv-E ⇒W_{1}=2.54-1.40=1.14 keV andW_{2}=2.54-2.02=0.52 keV**Q6.**The kinetic energy of an electron gets tripled, then the de-Broglie wavelength associated with it changes by a factor

Solution

de-Broglie wavelength of an electron is given by Î»=h/mv=h/√2mK Or Î»∝1/√K ∴ Î»

de-Broglie wavelength of an electron is given by Î»=h/mv=h/√2mK Or Î»∝1/√K ∴ Î»

^{'}/Î»=1/√3K √K/1=1/√3 Or Î»^{'}=Î»/√3 Hence, de-Broglie wavelength will change by factor 1/√3.**Q7.**In Milikan’s experiment, an oil drop having charge q gets stationary on applying a potential difference V in between two plates separated by a distance 'd

^{'}. The weight of the drop is

Solution

QE=mg⇒mg=QV/d

QE=mg⇒mg=QV/d

**Q8.**The de-Broglie wavelength of an electron, Î±- particle and proton all having the same kinetic energy is respectively given as Î»

_{e},Î»

_{Î±}and Î»

_{p}. Then which of the following is not true?

Solution

de-Broglie wavelength Î»=h/mv or Î»∝1/m ∴ Î»

de-Broglie wavelength Î»=h/mv or Î»∝1/m ∴ Î»

_{e}∝1/m, Î»_{Î±}∝ 1/m_{Î±}and Î»_{p}∝1/m_{p}As we know that m_{e}<m_{p}<m_{Î±}So, Î»_{e}>Î»_{p}>Î»_{Î±}Or Î»_{e}>Î»_{Î±}or Î»_{p}>Î»_{Î±}or Î»_{e}>Î»_{p}**Q9.**In the graph given below. If the slope is 4.12×10

^{-15}V-s, then value of 'h' should be

Solution

Slope of V

Slope of V

_{0}-v curve =h/e ⇒h=Slope ×e=1.6×10^{-19}×4.12×10^{-15}=6.6×10^{-34}J-s**Q10.**What is the strength of transverse magnetic field required to bend all the photoelectrons within a circle of a radius 50 cm. When light of wavelength 3800 â„« is incident on a barium emitter? (Given that work function of barium is 2.5 eV;h=6.63×10

^{-34}J-s; e=1.6×10

^{-19}C; m=9.1×10

^{-31}kg)

Solution

1/2 mv

1/2 mv

_{max}^{2}=[hc/Î»-Ï•_{0}] or v_{max}^{2}=2/m [hc/Î»-Ï•_{0}] =2/(9.1×10^{-31})×[(6.63×10^{-34}×3×10^{8})/(3.8×10^{-7})-2.5×1.6×10^{-19}] =27.12×10^{10}m^{2}s^{-2}Now, B=(mv_{max})/(eR_{max})=(9.1×10^{-31}×5.21×10^{5})/(1.6×10^{-19}×0.5) =6.32×10^{-6}T