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    (Original post by andyj72)
    sure, its T^2*k / 4п^2 = m

    u rearrange the formula, easy things can seem hard at times
    THANKS!!!!!!!!
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    (Original post by habosh)
    (T^2 * K)/(4 π²)=m
    THANKS!!!!!!
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    hi guys can someone help me here. Ive done the first part

    The graph shows how the intensity of
    light from a light-emitting diode (LED)
    varies with distance from the LED.

    Use data from the graph to show that the
    intensity obeys an inverse square law. {DONE}

    What does this suggest about the amount of
    light absorbed by the air? [3]

    is it because the light spreads out over a wider area therfore the light incident will decrease? it travels out radially?:confused:

    The light from the LED has a wavelength
    of 620 nm. Show that the energy of a
    photon of this light is approximately
    3 x 10-19 J. [2] {DONE}

    A student observes the LED from a
    distance of 0.20 m. The pupil of her eye
    has a diameter of 6.0 mm. Calculate the
    number of photons which enter her eye
    per second. [4] {HELPPPPPPPPPPPPPP}


    Explain in terms of photons why the light intensity decreases with increasing distance from the LED. [1]
    {HELP}
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    and another one soz guys

    When β radiation was first discovered, it was suggested that the atomic nucleus must contain electrons. However, it was soon realised that this was impossible because such electrons would possess far too much energy to be bound within the nucleus. Using the ideas of the earlier parts of this question, suggest why an electron confined within the nucleus would have a very high energy. [2]
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    (Original post by andyj72)
    hi guys can someone help me here. Ive done the first part

    The graph shows how the intensity of
    light from a light-emitting diode (LED)
    varies with distance from the LED.

    Use data from the graph to show that the
    intensity obeys an inverse square law. {DONE}

    What does this suggest about the amount of
    light absorbed by the air? [3]

    is it because the light spreads out over a wider area therfore the light incident will decrease? it travels out radially?:confused:

    The light from the LED has a wavelength
    of 620 nm. Show that the energy of a
    photon of this light is approximately
    3 x 10-19 J. [2] {DONE}

    A student observes the LED from a
    distance of 0.20 m. The pupil of her eye
    has a diameter of 6.0 mm. Calculate the
    number of photons which enter her eye
    per second. [4] {HELPPPPPPPPPPPPPP}


    Explain in terms of photons why the light intensity decreases with increasing distance from the LED. [1]
    {HELP}
    To use the data, you take measurements from the graph for the value of I and the corresponding value for r.

    use this I = k/r^2
    Use two corresponding measurements and rearrange k to make the subject both times, both times you will get the same value of k => Inverse square law is obeyed.

    Next bit..
    No light is absorbed by the air, as the value of k is constant throughout the operation.

    Last bit:
    A student observes the LED from a
    distance of 0.20 m. The pupil of her eye
    has a diameter of 6.0 mm. Calculate the
    number of photons which enter her eye
    per second. [4] {HELPPPPPPPPPPPPPP}

    use pir^2 to get the area. Power = intensity x area, use the value fors for intensity and area there.
    Therefore, the number of photons which enter her eye per second is given by ... The power calculated above / 3 x 10-19 J = Answer.
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    (Original post by andyj72)

    What does this suggest about the amount of
    light absorbed by the air? [3]

    is it because the light spreads out over a wider area therfore the light incident will decrease? it travels out radially?:confused:
    If the intensity obeys inverse square law, it means the light is not absorbed by the air on its way. Otherwise, it wouldn't obey inverse square law.
    The light from the LED has a wavelength
    of 620 nm. Show that the energy of a
    photon of this light is approximately
    3 x 10-19 J. [2] {DONE}

    A student observes the LED from a
    distance of 0.20 m. The pupil of her eye
    has a diameter of 6.0 mm. Calculate the
    number of photons which enter her eye
    per second. [4] {HELPPPPPPPPPPPPPP}
    Area of her eye, A = π*(0.003)²
    You should know the intensity, I at distance 0.20m from the graph(or calcuate it)
    Energy her eye receive in 1 second, P = I.A
    Then number of photons = P/energy of each photon(the value you got above)
    Explain in terms of photons why the light intensity decreases with increasing distance from the LED. [1]
    {HELP}
    Because photons spread out over a larger area as they travels longer, so number of photons reaching a futher point is less -> engery less -> intensity less.
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    (Original post by andyj72)
    and another one soz guys

    When β radiation was first discovered, it was suggested that the atomic nucleus must contain electrons. However, it was soon realised that this was impossible because such electrons would possess far too much energy to be bound within the nucleus. Using the ideas of the earlier parts of this question, suggest why an electron confined within the nucleus would have a very high energy. [2]
    wat r the earlier parts of the Q??

    Is it got 2 do with standing waves. The diameter of a nucleus is 1*10^-15

    Using de broglie's equation Lambda = h/p
    Then p(the electrons momentum will be p = h/Lambda)
    p = (6.63*10^-34)/(1*10^-15)
    p = 0.000000000000000000663
    p = mv
    therefore v = p/m
    v = = 0.000000000000000000663/9.11*10^-31 = 727771679473.11 m/s
    THats HUUGE...
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    thanx a million guys. i never though of any of this
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    (Original post by mackin boi)
    wat r the earlier parts of the Q??

    Is it got 2 do with standing waves. The diameter of a nucleus is 1*10^-15

    Using de broglie's equation Lambda = h/p
    Then p(the electrons momentum will be p = h/Lambda)
    p = (6.63*10^-34)/(1*10^-15)
    p = 0.000000000000000000663
    p = mv
    therefore v = p/m
    v = = 0.000000000000000000663/9.11*10^-31 = 727771679473.11 m/s
    THats HUUGE...

    yep it was to do with standing waves.
    the full question is below.

    . The electron in a hydrogen atom can be described by
    a stationary wave which is confined within the atom.
    This means that its de Broglie wavelength must be similar
    to the size of the atom, of the order of 10-10 m.
    Calculate the speed of an electron whose de Broglie
    wavelength is 1.0 x 10-10 m.
    Calculate the kinetic energy of this electron, in electron volts. [5]
    When β radiation was first discovered, it was suggested that the atomic nucleus must contain electrons. However, it was soon realised that this was impossible because such electrons would possess far too much energy to be bound within the nucleus. Using the ideas of the earlier parts of this question, suggest why an electron confined within the nucleus would have a very high energy. [2]
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    does anyone else find they can remember all these constants?
    I know all of them by heart, planck, mass and charge of an electron, G, 1/4piE0, etc...

    its funny I just realised I knew them all..
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    Whats the difference between permitivity & permability?

    Is permetitivity the 1/4piEo, in coloums law

    and permability something to do with magenatism?
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    the the (Unote) thingie ...in the magnetic flux density equations
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    (Original post by SinghFello)
    Whats the difference between permitivity & permability?

    Is permetitivity the 1/4piEo, in coloums law

    and permability something to do with magenatism?
    also used in capacitor delelectric i guess
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    (Original post by mik1w)
    does anyone else find they can remember all these constants?
    I know all of them by heart, planck, mass and charge of an electron, G, 1/4piE0, etc...

    its funny I just realised I knew them all..
    me 2...it's because when you use them so frequently you memorise them..shame we cant memorise the partical wave duality points and these stuff by heart
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    mackin boi, why are u considering diameter of d nucleus : 1* 10^-15 ?
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    (Original post by mikeA1)
    mackin boi, why are u considering diameter of d nucleus : 1* 10^-15 ?
    I think thats the actual constant your meant to know (Phy1). The diameter of an atom is 1*10^-10m. I think u have to know that aswell. Cud sum1 verify this.
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    (Original post by SinghFello)
    Whats the difference between permitivity & permability?

    Is permetitivity the 1/4piEo, in coloums law

    and permability something to do with magenatism?
    Permittivity is a constant that is in electric fields.
    Permeability is the same constant for magnetic fields.

    Trust me NO ONE can remember the difference. Its far simplier to say things like E-naught or mju-naught respectively.

    BTW
    E = q/(4pi e0 r^2)
    dB = mju0 dI ^ r/4pi r^3
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    (Original post by mackin boi)
    I think thats the actual constant your meant to know (Phy1). The diameter of an atom is 1*10^-10m. I think u have to know that aswell. Cud sum1 verify this.
    but should u be making use of that value for λ(lambda), shouldn't λ be tha value given( 1*10^-10)?
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    (Original post by mikeA1)
    but should u be making use of that value for λ(lambda), shouldn't λ be tha value given( 1*10^-10)?
    I think the Q says that the electron is in the nucleus which is 1*10^-15m...
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    what I'm tring 2 say is, u taking diameter to be lambda, should it be?
 
 
 
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