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Atoms Test - 37

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Atoms Test - 37
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  • Question 1
    1 / -0
    The property of cathode rays used in the monitor of a computer is :
    Solution

    A CRT monitor contains millions of tiny red, green, blue phosphor dots that glow when struck by an electron beam that travels across the screen to create visible image.

    This is called Fluorescence, this is shown in above diagram:


    In CRT tube cathode is heated filament. The heated filament is in a vacuum created inside a glass tube. The stream of electrons generated by an electron gun is powered out of cathode and is attracted to anode. This screen is coated     with phosphor, an organic material that glows when struck by electron beam. This process is called fluorescence.

  • Question 2
    1 / -0
    In a vacuum diode, some of the electrons possessing sufficient kinetic energy can reach the plate even if the plate potential is negative. For a certain diode, the plate current becomes zero when plate potential is $$-10  V$$. What is the kinetic energy of the electrons in $$eV$$, if the plate potential is $$-4  V$$ ?
    Solution
    $$\begin{aligned}&\begin{aligned}K E_{\max } &=e V_{\min }\\K E_{\max } &=\operatorname{10e} V\end{aligned}\\\text { KE, in this case }&=k E_{\max }-e V\\&\begin{aligned}=& \text { 10ev }-4 \mathrm{eV} \\=& 6\mathrm{eV}\end{aligned}\end{aligned}$$
  • Question 3
    1 / -0
    In a CRT the acceleration of electrons is controlled by changing ________ between the cathode and anode.
    Solution

    Positive potential is given to anode and negative potential is given to cathode. As a result of this potential difference, electrons generated by heating cathode are attracted towards anode due to this potential difference. If no potential difference is applied b/w anode and cathode in CRT, no electron would strike the screen. It is because of this potential difference b/w anode and cathode that the whole process occurs.

  • Question 4
    1 / -0
    The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about $$10^{-40}$$. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.
    Solution
    Radius of the first Bohr orbit is given by the relation,

    $$r_1=\dfrac{4\pi \epsilon_0(\dfrac{h}{2\pi})^2}{m_e e^2}$$      ......(1)

    Where,

    $$\epsilon_0=$$ Permittivity of free space

    $$h=$$ Planck's constant

    $$m_e=$$ Mass of an electron

    $$e=$$ Charge of an electron

    $$m_p=$$ Mass of a proton

    $$r=$$ Distance between the electron and the proton

    Coulomb attraction between an electron and a proton is given as:

    $$F_C=\dfrac{e^2}{4\pi \epsilon_0 r^2}$$     .....(2)

    Gravitational force of attraction between an electron and a proton is given as:

    $$F_G=\dfrac{Gm_p m_c}{r^2}$$     .....(3)

    Where

    $$G=$$ Gravitation constant $$=6.67 \times 10^{-11}\,Nm^2/kg^2$$

    If an electrostatic (Coulomb) force and the gravitation force between an electron and a proton are equal, then we can write:

    $$F_G=F_C$$

    $$\dfrac{Gm_p m_c}{r^2}=\dfrac{e^2}{4\pi \epsilon_0 r^2}$$

    $$\therefore \dfrac{e^2}{4\pi \epsilon_0}=Gm_p m_c$$     .....(4)

    Putting the value of equation ( 4 ) in equation ( 1 ), we get:

    $$r_1=\dfrac{(\dfrac{h}{2\pi})^2}{Gm_pm_c^2}$$

    $$=\dfrac{(\dfrac{6.63\times 10^{-34}}{2\times 3.14})^2}{6.67\times 10^{-11}\times 1.67 \times 10^{-27}\times (9.1\times 10^{-31})^2}\approx 1.21\times 10^{29}\,m$$


    It is known that the universe is $$156$$ billion light years wide or $$1.5\times 10^{27}\, m$$ wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.
  • Question 5
    1 / -0
    In a vacuum diode, $$2 \times {10}^{19}$$ electrons emitted from the cathode reach the plate of a vacuum diode in 4 seconds. Find the current and the power consumed if the potential difference between the cathode and the plate is $$200  V$$.
    Solution

    $$ I = \dfrac{q}{t} $$

    $$ =\dfrac { 2 \times 10^{19} \times 1.6 \times 10^{-19}}{4} $$

    $$ = 0.8 A $$

    $$ Power = VI $$

    $$ =200 \times 0.8 = 160 W $$

  • Question 6
    1 / -0
    The target used in a Coolidge tube for the production of X-rays is made up of ________ .
    Solution

    Molybdenum is used as a target in Coolidge tube for product.

    X- ray tube is an energy converter and it’s a device made of cathode and anode. An electrical current flows through the tube from cathode to anode,  when electron undergoes a energy loss, which results in generation of x -rays.

    Anode is a component in which x-rays are produced. It has 2 primary functions.

    1.To convert electricity to x-ray

    2.To dissipate heat in the process

    For this, material such as molybdenum and tungsten are used with high atomic number which has good heat storage capacity and low rate of evaporation.
  • Question 7
    1 / -0
    Who gave the Quantum model of hydrogen atom ?
    Solution
    Niels Bohr introduced the atomic Hydrogen model in 1913.
    Neils Bohr developed the Bohr model of the atom, in which he proposed that energy levels of electrons are discrete and that the electrons revolve in stable orbits around the atomic nucleus but can jump from one energy level (or orbit) to another.
  • Question 8
    1 / -0
    Gases exert pressure on the walls of the container, because the gas molecules
    Solution
    Gases exert pressure on the walls of the container, because the gas molecules possess momentum.
  • Question 9
    1 / -0
    How many protons and neutrons are there in the nuclei of $$^{25}_{12}Mg$$
    Solution
    Atomic no.(z)=12, mass no (A) = 25
    No of proton = Atomic no = 12
    No of neutron $$=$$ A $$-$$ no of proton
                          $$= 25-12 = 13$$
  • Question 10
    1 / -0
    Electromagnetic waves of wavelength ranging from  $$100\overset {\circ}{A}$$ to $$400 \overset {\circ}{A}$$ comes under :
    Solution
     NameWavelength Frequency (Hz) Photon Energy (eV) 
     GammaLess than 0.01 nm More than 10EHz 100kev - 300+ GeV 
     X-ray0.01-10 nm 30EHz-30 PHz 120 eV-120 keV 
    Ultraviolet 10 nm-400nm 30 Phz-790 THz 3 eV-124 ev 
    Visible  390nm-750nm 790THz-405Thz 1.7 eV-3.3eV 
    Infrared 750 nm -1mm 405 THz-300 GHz1.24meV-1.7eV 
    Microwave  1 mm- 1 meter  300 GHz -300MHz1.24 $$\mu $$eV-1.24 MeV 
    Radio 1 mm- km 300GHz -3Hz 12.4 feV-1.24 meV 

    The range  100 $$A^0$$ to 400$$A^0$$ come under Ultraviolet spectrum.
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