The figure shows stopping potential V-0 and frequency upsilon two different metallic surfaces A and B- The work of A- as compared to that of B is-morelessnothing can be saidequal

Question

Toppr Admin · Accepted Answer

The correct option is A less-xA0- -xA0- eV-h-x3C5-x2212-x3D5-xA0-xA0-From above equation-The value of threshold frequency -x3BD-0 for A is less than that for B - hence -x3D5-A -lt- -x3D5-B-

The figure shows stopping potential $V_{0}$ and frequency $υ$ for two different metallic surfaces A and B. The work function of A, as compared to that of B is:

more
less
nothing can be said
equal

The correct option is A less
$e V = h υ - ϕ$
From above equation,
The value of threshold frequency $ν_{0}$ for A is less than that for B , hence $ϕ_{A}$ < $ϕ_{B}$ .

The figure shows stopping potential V0 and frequency υ for two different metallic surfaces A and B. The work function of A, as compared to that of B is:morelessnothing can be saidequal

The correct option is A less eV=hυ−ϕ From above equation,The value of threshold frequency ν0 for A is less than that for B , hence ϕA < ϕB.

The figure shows stopping potential $V_{0}$ and frequency $υ$ for two different metallic surfaces A and B. The work function of A, as compared to that of B is:

more
less
nothing can be said
equal

The correct option is A less
$e V = h υ - ϕ$
From above equation,
The value of threshold frequency $ν_{0}$ for A is less than that for B , hence $ϕ_{A}$ < $ϕ_{B}$ .