(Round off to the Nearest Integer).
[Use : $\sqrt{3}=1.73, h =6.63 \times 10^{-34} Js$ $m _{ e }=9.1 \times 10^{-31} kg ; c =3.0 \times 10^{8} ms ^{-1}$ $\left.1 eV =1.6 \times 10^{-19} J \right]$
$=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{248 \times 10^{-9} \times 1.6 \times 10^{-19}} eV$
$=\frac{6.63 \times 3 \times 100}{248 \times 1.6}$
$=0.05 eV \times 100=5 eV$
Now using $E =\phi+ K . E$
$5=3+ K . E$
$K . E .=2 eV =3.2 \times 10^{-19} J$
for debroglie wavelength $\lambda=\frac{ h }{ mv }$
$K . E =\frac{1}{2} mv ^{2}$
so $v =\sqrt{\frac{2 KE }{ m }}$
hence $\lambda=\frac{ h }{\sqrt{2 KE \times m }}$
$=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 3.2 \times 10^{-19} \times 9.1 \times 10^{-31}}}$
$=\frac{6.63}{7.6} \times \frac{10^{-34}}{10^{-25}}=\frac{66.3 \times 10^{-10} m }{7.6}$
$=8.72 \times 10^{-10} m$
$\approx 9 \times 10^{-10} m$
$=9 ^o$
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$C_2H_4(g) + H_2(g) \to C_2H_6(g)$
.......$kJ$
| Bond | Bond energy $(kJ)$ |
| $C-H$ |
$413$ |
| $C-C$ | $348$ |
| $C=C$ | $610$ |
| $H-H$ | $436$ |
| Column $I$ | Column $II$ |
| $(A)$ Silica gel | $(i)$ Transistor |
| $(B)$ Silicon | $(ii)$ Ion-exchanger |
| $(C)$ Silicone | $(iii)$ Drying agent |
| $(D)$ Silicate | $(iv)$ Sealant |