A student performs an experiment to observe photoelectric effect by shining light on a metal surface- The light source limits light of wavelength 450 nm - The table given below lists the available metals and their corresponding work functions beginarray -WhineMetal - -phiev IIVhline Tungsten - 4-5 INhline Lithium - 2-3 -llhline Barium - 2-5 lIVhline Tantalum - 4-2 Whlinearray Which of the above metals can be used to produce electrons by the photoelectric effect from the given source of light-A- Barium onlyB- Tungsten and lithium onlyC- Barium or LithiumD- Lithium- Tantalum or tungsten

The correct option is C Barium or Lithium Energy of incident photon =1240ev nm/λ_(nm) = ( 1240/450 )eV = 2.75eV For photoelectric effect, E_incident gt; ϕ ...

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Standard XII

Chemistry

Reactivity with Air CO2 and Water (Group 1)

A student per...

Question

A student performs an experiment to observe photoelectric effect by shining light on a metal surface. The light source limits light of wavelength 450nm. The table given below lists the available metals and their corresponding work functions
$\begin{matrix} M e t a l & ϕ (e v) T u n g s t e n & 4.5 L i t h i u m & 2.3 B a r i u m & 2.5 T a n t a l u m & 4.2 \end{matrix}$

Which of the above metals can be used to produce electrons by the photoelectric effect from the given source of light?

Barium only

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Barium or Lithium

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Lithium, Tantalum or tungsten

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Tungsten and lithium only

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Solution

The correct option is B
Barium or Lithium

Energy of incident photon $= \frac{1240 e v - n m}{λ_{(} n m)}$
$= (\frac{1240}{450}) e V = 2.75 e V$
For photoelectric effect,
$E_{i n c i d e n t} > ϕ$

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Q. A physicist wishes to eject electrons by shining light on a metal surface. The light source emits light of wavelength of 450 nm. The table lists the only available metal and their work functions.

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A physicist wishes to eject electrons by shining light on a metal surface. The light source emits light of wavelength of 450 nm. The table lists the only available metal and their work functions.

Q. Which metal(s) can be used to produce electrons by the photoelectric effect from given source of light ?

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The correct option is B Barium or Lithium Energy of incident photon =1240ev−nmλ(nm) =(1240450)eV=2.75eV For photoelectric effect, Eincident>ϕ

The correct option is B
Barium or Lithium

Energy of incident photon $= \frac{1240 e v - n m}{λ_{(} n m)}$
$= (\frac{1240}{450}) e V = 2.75 e V$
For photoelectric effect,
$E_{i n c i d e n t} > ϕ$