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26.4k views 4.9 unacademy Kumar Ketan #16 Educator in IIT JEE #6 years teaching experience #IT JEE Physics #YouTuber (150k+ subs) #Unacademy educator(25kt lifetimeviews) #Under graduate topper Follow me on the Unacademy Learning App 1.3k 2 26 Followers Following Courses Get updates about new courses Watch all my lessons .Download slides and watch offline Lists (1) Kumar Ketan HINDI IIT JEE Physics by Kumar Ketan for lIT JEE 8 4 saves Kumar Ketan
#200 Most Important Questions of Modern Physics for IJT JEE MAINS and ADVANCE Electron Beam Dissociative Result Fiber 2e. Precursor Mirror Charged and Neutral Fragments Light source Non-dissociative Result ivot Parent Molecule Scale lonized Parent Molecule
#Q.28 If E,,E, and E, respectively the kinetic energies of an electron, an a -particle and a proton each having same de-Broglie wavelength, then b) E2>E,>E
V2Em where , h and E are constant. So,
#Q.29 m, and m2 having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles 2 is (a) mj/ m2 Solution From the law of conservation of momentum, A particle of mass M at rest decays into two particles of masses (JEE 1999) (b) m2/mi 2'Vm ( in opposite directions) Now, de-Broglie wavelength is given b --, where h Planck constant Since magnitude of momentum(p) of both the particles is equal, therefore - 2 or Therefore, the correct option is (c).
#Q.30 According to Einstein's photoelectric equation, the plot of the maximum kinetic energy of the emitted photoelectrons from a metal versus frequency of the incident radiation gives a straight line whose slope a) Depends on the nature of material used b) Depends on the intensity of radiation c) Depends on both intensity of radiation and the nature of metal used d) Is the same for all metals and independents of the intensity of radiation
Soln. Kmax = hf-W Kmax versus f graph is a straight line of slope h (a universal constant)
#Q.31 The velocity of the electron in the first Bohr orbit as compared to that of light is about a) 1/300 c) 1/137 b) 1/500 d) 1/187
Soln y'=2.19 106m/s ~ .c 137 1 (1.6 10-19)2 Ze2 2Eonh 2 8.85 10-12 1 6.6 1034
Soln. Let -particles are n and -particles are m Then, 86-2n+m-84 222-4n-210 solving these two equations . we get n-3 andp4
Soin. 1min = V(involts) 12375 . 0 in A 20x10 =0.62