Question

2. In an experiment on photoelectric effect, stopping potential is 1.0 V when
light of wavelength 6520 Å is incident on the emitting surface. The work
function of the metal is
(i) 0.9 eV
(iii) 1.9 eV
1.45 eV
(iv) 5.8 eV​

Answer 1

Answer:

The maximum kinetic energy for the photoelectrons is 

Emax=hν−ϕ

where, ν is the frequency of incident light and ϕ is photoelectric work function of metal.

If Vo is the stopping potential then

eV0=hλc−ϕ .....................(since, ν=λc)

As per the problem, for incident light of 4000A∘, the stopping potential is 2V. When the wavelength of incident light is reduced to 3000A∘, then the stopping potential will increase to value more than 2V(as per the above equation).

So, the answer is option (B).

Answer 2

Answer:

The work function of the metal is $W=0.9 e V$

Explanation:

Absolute approaches allow for direct measurement of the work function value. The electrons in the metal are given enough kinetic energy to overcome the barrier at the metal/vacuum contact, allowing them to escape the metal and obtaining the work function from the ensuing electric current.

$\begin{aligned}&\frac{h c}{\lambda}=W+e V_{s} \\&\frac{h c}{652 n m}=W+1 e V \\&\text { Also, } \frac{h c}{326 n m}=W+2.9 e V\end{aligned}$

Equate  form both side

$\begin{aligned}&652(W+1)=326(W+2.9) \\&2(W+1)=W+2.9 \\&W=2.9-2 \\&W=0.9 \mathrm{eV}\end{aligned}$

Therefore the work function of the metal is $W=0.9 e V$