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11.30 Light of intensity 10–5 W m–2 falls on a sodium photo-cell of surface area 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?

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    Answered by

    Payal Gupta | Contributor-Level 10

    5 months ago

    11.30 Intensity of the incident light, I = 10 - 5  W m - 2

    Surface area of the sodium photocell, A = 2 c m 2  = 2 × 10 - 4 m 2

    The incident power of the light, P = I × A  = 10 - 5 × 10 - 4 m 2 W = 2 × 10 - 9  W

    Work function of the metal, ∅ 0 = 2 eV = 2 × 1.6 × 10 - 19  J = 3.2 × 10 - 19  J

    Number of layers of sodium that absorbs the incident energy, n = 5

    The effective atomic area of a sodium atom, A e  = 10 - 20 m 2

    Hence, the number of conduction electrons in n layers is given by:

    n’ = n × A A e  = 5 × 2 × 10 - 4 10 - 20  = 1 × 10 17

    Since the incident power is uniformly absorbed by all the electrons continuously, the amount of energy absorbed per second per electron is

    E = P n '  = 2 × 10 - 9 1 × 10 17  = 2 × 10 - 26  J/

    ...more

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Given below are two statements:

Statement I : Davission-Germer experiment establishes, the wave nature of electrons.

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Radiation from hydrogen gas excited to first excited state is used for illuminating certain metallic plate. When the same plate is exposed to the radiation from some unknown hydrogen like gas excited to the same level it is found that de-Broglie wavelength of fastest photoelectron has decreased by 2.3 times. It is given that energy corresponding to longest wavelength of Lyman series of the unknown gas is 3 times the ionization energy of hydrogen gas (13.6eV). The work function (in eV) of the metallic plate is ______. [Take (2.3)2 = 5.25]
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z² × (13.6) (1 - ¼) = 3 × (13.6)
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  K . E 1 = 1 2 3 0 8 0 0 − ? - (i)

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When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is υ 1 . When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes υ 2 . If υ 2 = x υ 1 , the value of x will be __________.             
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Kindly consider the following

f(x)={2x2−3x+2x−2,x≠25,x=2
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This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

f(x)=2x2−3x−2x−2                                      =2x2−4x+x−2x−2=2x(x−2)+1(x−2)x−2                                       =(2x+1)(x−2)x−2=2x+1                           limx→2−f(x)=2x+1=limh→02(2−h)+1=4+1=5                          limx→2+f(x)=2x+1=limh→02(2+h)+1=4+1=5                           limx→2f(x)=5As   limx→2−f(x)=limx→2+f(x)=limx→2f(x)=5Hence,  f(x)  is  continuous  at  x=2.

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