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The threshold frequency for a certain metal is $3.3 \times 1014Hz$. If light of frequency $8.2 \times 1014 Hz$ is incident on the metal, predict the cutoff voltage for the photoelectric emission.

$\begin {array} {1 1} (a)\;20.292V & \quad (b)\;20292 \times 10^{-4}kV \\ (c)\;202.92V & \quad (d)\;2.0292V \end {array}$

 

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Ans : (d) 2.0292V
$v_{\circ} = 3.3 \times 10^{14}Hz\: \: \: \: \:v=8.2 \times 10^{14}Hz$
cut off voltage $= V_{\circ}$
So, $eV_{\circ} = h(v-v_{\circ})$
$ \Rightarrow V_{\circ} = h(v – v_{\circ})/e$
$= 6.626 \times 10^{-14} \times (8.2 \times 10^{14} – 3.3 \times 10^{14} ) / 1.6 \times 10^{19}$
$= 2.0292 V$

 

answered Dec 23, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

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