Home
Class 12
PHYSICS
The work function of sodium is 2.3 eV . ...

The work function of sodium is 2.3 eV . The threshold wavelength of sodium will be

A

2900 Å

B

2500 Å

C

5380 Å

D

2000 Å

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Topper's Solved these Questions

  • ELECTRON, PHOTON, PHOTOELECTRIC EFFECT AND X-RAYS

    ERRORLESS |Exercise Cathode Rays and Positive Rays|11 Videos

  • ELECTRON, PHOTON, PHOTOELECTRIC EFFECT AND X-RAYS

    ERRORLESS |Exercise Photon and Photoelectric Effect|4 Videos

  • ELECTRO MAGNETIC INDUCTION

    ERRORLESS |Exercise SET|20 Videos

  • ELECTRONICS

    ERRORLESS |Exercise Selv Evaluation Test|23 Videos

Similar Questions

Explore conceptually related problems

The work function of cesium is 2.14 eV. The threshold freqency of caesium is

The work function of Na metal is 2.3 eV , then the threshold wavelength lies in the following region of EM spectrum -

The work function of a metal is 4.2 eV , its threshold wavelength will be

If the work function of a photo - metal is 6.825 eV . Its threshold wavelength will be ( c = 3 xx 10^(8) m//s)

Light strikes a sodium surface, causing photoelectric emission. The stopping potential for the ejected electrons is 5.0 V, and the work function of sodium is 2.2 eV. What is the wavelength of the incident light?

The work function for a metal whose threshold wavelength is 5000 Å is

If the work function of a and is 3 eV, calculate the threshold wavelength of that metal. (Velocity of light =3xx10^(8)m//s Planck's constant= 6.63xx10^(-34)J-s,1eV=1.6xx10^(-19) )

The work function for sodium metal is 2.46eV . Determine the cutoff wavelength for sodium? Strategy: The cutoff frequency v_(0) is related to the work function through the relation v_(0) = W_(0)//h . This corresponds to a cutoff wavelength of lambda_(0) = (C )/(v_(0)) = (C)/(W_(0)//h) = (hc)/(W_(0)) . wavelength grater than lambda_(0) for a metal with work function W_(0) produce no photoelectric effect.