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 Illhline Lithium - 2-3 -Ilhline Barium - 2-5 lIVhlineTantalum - 4-2 illhline end array Suppose the photoelectric experiment is done separately with these metals with light of wavelength 450 nm- The maximum magnitude of stopping potential amongst all the metals isA- 2-75 VB- 0-25 VC- 4-5 VD- 0-45 V

The correct option is D 0.45 V K_max = eV_0 ⇒ V_0 = K_max/e = E_incident ϕ/e = ( 2.75 2.3/e ) = 0.45V ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Intensity in Photon Model

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
$\begin{matrix} M e t a l & ϕ (e v) T u n g s t e n & 4.5 L i t h i u m & 2.3 B a r i u m & 2.5 T a n t a l u m & 4.2 \end{matrix}$

Suppose the photoelectric experiment is done separately with these metals with light of wavelength 450nm. The maximum magnitude of stopping potential amongst all the metals is

2.75 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.5 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.45 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.25 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
0.45 V

$K_{m a x} = e V_{0}$
$\Rightarrow V_{0} = \frac{K_{m a x}}{e} = \frac{E_{i n c i d e n t} - ϕ}{e} = (\frac{2.75 - 2.3}{e}) = 0.45 V$

Suggest Corrections

Similar questions

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?

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.

\begin{matrix} M e t a l & W_{0} (e V) B a r i u m & 2.5 L i t h i u m & 2.3 T a n t a l u m & 4.2 T u n g s t e n & 4.5 \end{matrix}

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

Q. Suppose photoelectric experiment is done separately with these metals with light of wavelength 450 nm. The maximum magnitude of stopping potential amongst all the metals is ______________.

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.
$\begin{matrix} M e t a l & W_{0} (e V) B a r i u m & 2.5 L i t h i u m & 2.3 T a n t a l u m & 4.2 T u n g s t e n & 4.5 \end{matrix}$

Which option correctly identifies the metal that will produce the most energetic electrons and their energies?

Which option correctly identifies the metal that will produce the most energetic electrons and their energies?

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.

Join BYJU'S Learning Program

The correct option is C 0.45 V Kmax=eV0 ⇒V0=Kmaxe=Eincident−ϕe=(2.75−2.3e)=0.45V

The correct option is C
0.45 V

$K_{m a x} = e V_{0}$
$\Rightarrow V_{0} = \frac{K_{m a x}}{e} = \frac{E_{i n c i d e n t} - ϕ}{e} = (\frac{2.75 - 2.3}{e}) = 0.45 V$