Question

A certain metal when irradiated by light $(r=3.2\times {10}^{16}Hz)$ emits photo electrons with twice kinetic energy as did photo electrons when the same metal is irradiated by light $(r=2.0\times {10}^{16}Hz)$. The $v_0$ of metal is:

• A. $1.2\times {10}^{14}Hz$
• B. $8\times {10}^{15}Hz$
• C. $1.2\times {10}^{16}Hz$
• D. $4\times {10}^{12}Hz$

Answer 1

The correct option is B $8\times {10}^{15}Hz$

${\left(KE\right)}_1=h{v}_1-h{v}_0$
${\left(KE\right)}_2=h{v}_2-h{v}_0$
As, ${\left(KE\right)}_1=2\times {\left(KE\right)}_2$
$\therefore h{v}_1-h{v}_0=2(h{v}_2-h{v}_0)$
or, $h{v}_0=2h{v}_2-h{v}_1$
or, $v_0=2v_2-v_1$
$=2\times (2\times {10}^{16})-(3.2\times {10}^{16})$
$=0.8\times {10}^{16}Hz=8\times {10}^{15}Hz$