The correct order of different types of energies is
(a) Eel >> Evib >> Erot >> Etr
(b) Eel >> Erot >> Evib >> Etr
(c) Eel >> Evib >> Etr >> Erot
(d) Etr >> Evib >> Erot >> Eel

Some questions (Assertion-Reason Type) are given below. Each question contains Statement I (Assertion) and statement II(reason). Each question has 4 choices (a),(b),(c ) and (d) out of which only one is correct. So select the correct choise. a. Statement I is True, Statement II is True,Statement II is a correct explanation for Statement I b. Statement I is True, Statement II is True, Statement II is NOT a correct ecplanation for Statement I c. Statement I is True, Statement II is False . d. Statement I is false, Statement II is True. 2. Statement:I Though light of a single frequency (monochromatic light) is incident on a metal, the energies of emitted photoelectrons are different. Statement II: The energy of electrons just after they absorb photons incident on the metal surface may be lost in collision with other atoms in the metal before the electron is ejected out of the metal.

Following experiment was performed by J.J. Thomson in order to measure ratio of charge e and mass m of electron. Electrons emitted from a hot filament are accelerated by a potential difference V. As the electrons pass through deflecting plates, they encounter both electric and magnetic fields. the entire region in which electrons leave the plates they enters a field free region that extends to fluorescent screen. The entire region in which electrons travel is evacuated. Firstly, electric and magnetic fields were made zero and position of undeflected electron beam on the screen was noted. The electric field was turned on and resulting deflection was noted. Deflection is given by d_(1) = (eEL^(2))/(2mV^(2)) where L = length of deflecting plate and v = speed of electron. In second part of experiment, magnetic field was adjusted so as to exactly cancel the electric force leaving the electron beam undeflected. This gives eE = evB . Using expression for d_(1) we can find out (e)/(m) = (2d_(1)E)/(B^(2)L^(2)) If the electron speed were doubled by increasing the potential differece V, which of the following would be true in order to correctly measure e//m

Following experiment was performed by J.J. Thomson in order to measure ratio of charge e and mass m of electron. Electrons emitted from a hot filament are accelerated by a potential difference V. As the electrons pass through deflecting plates, they encounter both electric and magnetic fields. the entire region in which electrons leave the plates they enters a field free region that extends to fluorescent screen. The entire region in which electrons travel is evacuated. Firstly, electric and magnetic fields were made zero and position of undeflected electron beam on the screen was noted. The electric field was turned on and resulting deflection was noted. Deflection is given by d_(1) = (eEL^(2))/(2mV^(2)) where L = length of deflecting plate and v = speed of electron. In second part of experiment, magnetic field was adjusted so as to exactly cancel the electric force leaving the electron beam undeflected. This gives eE = evB . Using expression for d_(1) we can find out (e)/(m) = (2d_(1)E)/(B^(2)L^(2)) A beam of electron with velocity 3 xx 10^(7) m s^(-1) is deflected 2 mm while passing through 10 cm in an electric field of 1800 V//m perpendicular to its path. e//m for electron is

Following experiment was performed by J.J. Thomson in order to measure ratio of charge e and mass m of electron. Electrons emitted from a hot filament are accelerated by a potential difference V. As the electrons pass through deflecting plates, they encounter both electric and magnetic fields. the entire region in which electrons leave the plates they enters a field free region that extends to fluorescent screen. The entire region in which electrons travel is evacuated. Firstly, electric and magnetic fields were made zero and position of undeflected electron beam on the screen was noted. The electric field was turned on and resulting deflection was noted. Deflection is given by d_(1) = (eEL^(2))/(2mV^(2)) where L = length of deflecting plate and v = speed of electron. In second part of experiment, magnetic field was adjusted so as to exactly cancel the electric force leaving the electron beam undeflected. This gives eE = evB. Using expression for d_(1) we can find out (e)/(m) = (2d_(1)E)/(B^(2)L^(2)) If the electron is deflected downward when only electric field is turned on, in what direction do the electric and magnetic fields point in second part of experiment

Consider the following statements, A. Colour of a transition metal complex is dependent on the nature of energy difference between two d-levels. B. colour of the complex is dependent on the nature of the ligand and the type of complex formed. C. ZnSO_(4) and TiO_(2) white and in both d-dspectra are impossible. Select the correct statements.

A laser (light amplification by stimulated emission of radition ) provides a coherent (in phase) monochoramtic source of high intensity light . Lasers are used in eye surgery,CD/DV players basic research etc. Some modern eye lasers can be "tuned" to emit a desired wavelenght . The following table shows the lambda=(nm),v(s^(-1)) and E(J) but some of the date missing : [Take E(ev)=(1240)/lambda_(nm)] {:("Leser No.",lambda("nm"),v(s^(-1)),"Energy (J)"),("I",A,5xx10^(-14),B),("II",487.9,C,D),("III",E,F,3.2xx10^(-19)):} Which of the following statements(s) is /are correct? (i) All the three radiation (from laser ) lie in visble region (ii) The enegies D and B alone are not sufficient to remove electron completely from isolated gaseous H-atoms which is present at ground state (iii) In the spectrum the increasing order of lambda for the type of lasers is IIltIIIltI (iv) None of the above statement is correct