Question 173: Dual Nature of Matter and Radiation

Question

The stopping potential V
0
(in volt) as a function of frequency ($\upsilon $) for a sodium emitter, is shown in the figure.
The work function of sodium, from the data plotted in the figure, will be:
(Given: Planck’s constant (h) = 6.63 × 10
–34
Js, electron charges e = 1.6 × 10
–19
C)
![](https://imagex.cdn.examgoal.net/7kejek1by3qdu/4d299145-faad-4725-89cf-c8254010e8ad/090a2320-e68c-11e9-a7bc-b7b08ce8d541/file-7kejek1by3qdv-257w.jpg)

Accepted Answer

$hv = \varphi  + e{v_0}$
${v_0} = {{hv} \over e} - {\varphi  \over e}$
v
0
is zero for v = 4 ÃÂ 10
14
Hz
$0 = {{hv} \over e} - {\varphi  \over e}$
$ \Rightarrow \phi  = hv$
$ = {{6.63 	imes {{10}^{ - 34}} 	imes 4 	imes {{10}^{14}}} \over {1.6 	imes {{10}^{ - 19}}}} = 1.66\,eV$

Answer

1.95 eV

Answer

2.12 eV

Answer

1.82 eV

Answer

1.66 eV

Dual Nature of Matter and Radiation

The stopping potential V 0 (in volt) as a function of frequency (\upsilon ) for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be: (Given: Planck’s constant (h) = 6.63 × 10 –34 Js, electron charges e = 1.6 × 10 –19 C)

Choose the correct answer: