The energy equivalent to `10eV` is
O `1.6times10^(-19)J`
O `1.6times10^(-12)J`
O `1.6times10^(-13)J`
O `1.6times10^(-18)J`

A 15pF capacitor is connected to a 60V battery.How much electrostatic energy is stored? (A) 2.70times10^(-8)J (B) 1.77times10^(-8)J (C) 2.35times10^(-8)J (D) 1.35times10^(-8)J

The average thermal energy of oxygen at a temperature of 27C (А) 6.2times10^(-21) J (В) 4.5times10^(-21) J (C) 3.5times10^(-21) J (D) 1.2times10^(-21) J

The charge equivalent to 6times10^(18) electrons is .

What will be the angular momentum of a electron,if energy of this electron in H-atom is 1.5eV (in J-sec) (A) 1.05times10^(-34) (B) 2.1times10^(-34) (C) 3.15times10^(-34) (D)-2.1times10^(-34)

Simplify: 10^(-1)times10^(2)times10^(-3)times10^(4)times10^(-5)times10^(6)

Calculate the wavelength of an electron that has been accelerated in a particle accelerator through a potentiation difference of 100 million volts [ 1eV = 1.6 xx 10^(-19)J, m_(e) = 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) J s, c = 3.0 xx 10^(8) m s^(-1)]

The correct expanded form of "9.06" is (a) (9times10)+(6times(1)/(10)) , (b) (9times1)+(6times(1)/(10)) (c) (9times1)+(6times(1)/(100)) , (d) (9times10)+(6times(1)/(100))

The operating magnetic field for accelerating protons in a cyclotron oscillator having frequency of 12 MHz is (q= 1.6 xx 10^(-19) C, m_(p) = 1.67 xx 10^(-27) kg and 1 MeV = 1.6 xx 10^(-13J))

An electron beam energes from an accelerator with kinetic energy 100eV. What is its de-Broglie wavelength? [m = 9.1 xx 10^(-31)kg, h = 6.6 xx 10^(-34)Js, 1eV = 1.6 xx 10^(-19)J]

Photon having energy equivalent to the binding energy of 4th state of He^(+) ion is used to eject an electron from the metal with KE/2eV . If electron is further accelerated through a potential difference of 4V then the minimum value of de Broglie wavelength associated with the electron is : (h = 6.6 xx 10^(-34) J-s , m_(e) = 9.1 xx 10^(-31) kg . 1 eV = 1.6 xx 10^(-19) J )