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Electrons are emitted by a hot filament ...

Electrons are emitted by a hot filament and are accelerated by an electric field as shown in figure .The two stops at the left ensure that the electron beam has a uniform cross-section.

A

(a)The speed of the electron is more at B than at A.

B

(b)The electric current is from left to right.

C

(c )The magnitude of the current is larger at B than at A .

D

(d)The current density is more at B than at A.

Text Solution

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The correct Answer is:
A
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Electron are emited from an electron gun at almost zero velocity and are accelerated by an electric field E through a distance of 1.0m The electron are now scatteared by an atomic hydrogen sample in ground state what should be the minimum value of E so that red light of wavelength 656.5 nm may be emitted by the hydrogen?

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Knowledge Check

  • The de Broglie wavelength of an electron accelerated by an electric field of V volt is given by:

    A
    `lamda=(1.23)/(sqrt(m))`
    B
    `lamda=(1.23m)/(sqrt(h))`
    C
    `(1.23)/(sqrt(V))nm`
    D
    `lamda=(1.23)/(V)`.

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    A
    the speed of the electrons increases
    B
    the current constituted by the beam increases
    C
    the number of electrons per unit volume in the beam increases
    D
    the number of electrons per unit volume in the beam decreases

  • 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

    A
    `1.5 xx 10^(11)Ckg^(-1)`
    B
    `2 xx 10^(11)Ckg^(-1)`
    C
    `2.5 xx 10^(11)Ckg^(-1)`
    D
    `3 xx 10^(11)Ckg^(-1)`

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    Consider one design of electron gun, which has a barrel of length 1 m made of an insulating material. At one end of the barrel, a filament is electrically heated so that it emits electrons. Assume these electrons are emitted with a the sides negligible speed. What electric field should be applied parallel to the barrel so that electrons may achieve speed of light before leaving the barrel? Ignore the variation of mass of electron due to its speed.

    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

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