Question

the threshold frequency for a metal is 5×10^14/s what will be kinetic energy of a photoelectron emitted when radiation of frequency v=1.5×10^15/s strikes on a metal surface. (1)h×10^14 J (2)h×10^16 J (3)h×10^16 J (4)h J​

Answer 1

Answer:

  K.E.= 1/2meV2= h(v−v0)

∴ K.E .=(6.626×10^−34) (1.1×10^15−6×10^14)

∴ K.E.=  (6.626×10^−34) (5×10^14)

        =  3. 313 × 10^−20J

Answer 2

Answer:

The kinetic energy of the emitted photoelectron is $h\times 10 ^{15}$ J.

Explanation:

Given,

The threshold frequency for a metal, $\nu\textdegree$ = $5\times 10^{14}s^{-1}$

The light's frequency, $\nu$ = $1.5\times 10^{15}s^{-1}$

The kinetic energy of the emitted photoelectron =?

As we know,

• According to the photoelectric effect;
• K.E = $h\nu - h\nu\textdegree$

Here, h is Planck's constant.

After putting the values in the equation, we get:

• K.E = $h( 1.5 \times 10^{15}) -h ( 5\times 10^{14} )$
• K.E = $h \times 10^{14} [ 1.5\times 10 - 5]$
• K.E = $h\times 10^{14} [ 15-5]$
• K.E = $h\times 10^{15}$ J.

Hence, the kinetic energy of the emitted photoelectron = $h\times 10 ^{15}$ J.