Home
Class 12
PHYSICS
A metal is illumimated by light of two d...

A metal is illumimated by light of two different wavelength `248nm` and `310 nm` . The maximum speeds of the photoelectrons corresponding in these wavelength are `u_(1) and u_(2)` respectively . If the ratio `u_(1) : u_(2) = 2 : 1` and `hc = 1240 eVnm `, the work function of the meal is nearly

A

3.7 eV

B

3.2 eV

C

2.8 eV

D

2.5 eV

Text Solution

Verified by Experts

The correct Answer is:
A

`K_("max")=(hc)/lambda -phi_(0)`
`(K_(max_(1)))/(K_(max_(2)))=4/1 =((1240)/248-phi_(0))/((1240)/310-phi_(0))`
`rArr phi_(0) = 3.7 eV `
Promotional Banner

Similar Questions

Explore conceptually related problems

A metal is illumimated by light of two different wavelength 248nm and 310 nm . The maximum speeds of the photoelecrtron corresponding in these wavwlength are u_(1) and u_(2) respectively . If the ratio u_(1) : u_(2) = 2 : 1 and hc = 1240 eVnm , the work function of the inetial is rearly

A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are mu_1 and mu_2 respectively. If the ratio u_1: u_2 = 2:1 and hc= 1240 eV nm, the work function of the metal is neraly. (a)3.7 eV (b) 3.2 eV (c ) 2.8eV (d) 2.5eV.

When a metal surface is illuminated by light wavelengths 400 nm and 250 nm , the maximum velocities of the photoelectrons ejected are upsilon and 2 v respectively . The work function of the metal is ( h = "Planck's constant" , c = "velocity of light in air" )

Surface of certain metal is first illuminated with light of wavelength lambda_(1)=350 nm and then, by light of wavelength lambda_(2)=540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to : (Energy of photon =(1240)/(lambda("in nm"))Ev

A photonsensitive metallic surface is illuminated alternately with light of wavelength 3100 Å and 6200 Å . It is observed that maximum speeds of the photoelectrons in two cases are in ratio 2 : 1 . The work function of the metal is (hc = 12400 e VÅ)

The surface of a metal is illuminated with light of wavelength 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. What is the work function of the metal ? [ hC=1240 eV*nm ]

Trajectories of two projectiles are shown in the figure. Let T_(1) and T_(2) be the time period and u_(1) & u_(2) speeds of projection. Then

Lights of wavelenths lambda_(1)=340nm and lambda_(2)=540nm are incident on a metallic surface. If the ratio of the maximum speeds of electrons ejected is 2, the work function of the metal is