The maximum kinetic energy of the photoelectrons ejected from a metal when it is irradiated with a radiation of frequency 2- 10-14 s-1 is 6-63 - 10-20 J- The thereshold frequency of the metal is-2-10-14-s-11-10-14-s-12-10-14-s-13-10-14-s-1

The correct option is B 1 ×10^14 s^ 1hv=hv_o +KE hv_o=hv KE 6.626 × 10^ 34× v_o=6.626 × 10^ 34×2× 10^14 6.63× 10^ 20 v_o=1.3252×10^ 19 6.63×10^ 20/6.626×10^ ...

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Standard XII

Chemistry

Emission and Absorption Spectra

The maximum k...

Question

The maximum kinetic energy of the photoelectrons ejected from a metal when it is irradiated with a radiation of frequency $2 \times 10^{14} s^{- 1}$ is $6.63 \times 10^{- 20} J$ . The thereshold frequency of the metal is:

2 \times 10^{14} s^{- 1}

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3 \times 10^{14} s^{- 1}

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2 \times 10^{- 14} s^{- 1}

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1 \times 10^{14} s^{- 1}

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Solution

The correct option is D $1 \times 10^{14} s^{- 1}$
$h v = h v_{o} + K E$
$h v_{o} = h v - K E 6.626 \times 10^{- 34} \times v_{o} = 6.626 \times 10^{- 34} \times 2 \times 10^{14} - 6.63 \times 10^{- 20} v_{o} = \frac{1.3252 \times 10^{- 19} - 6.63 \times 10^{- 20}}{6.626 \times 10^{- 34}} = 9.99 \times 10^{13} = 10^{14} s^{- 1}$

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The maximum kinetic energy of the photoelectrons ejected from a metal when it is irradiated with a radiation of frequency 2×1014s−1 is 6.63×10−20J. The thereshold frequency of the metal is:

The correct option is D 1 ×1014 s−1hv=hvo+KE hvo=hv−KE6.626×10−34×vo=6.626×10−34×2×1014−6.63×10−20vo=1.3252×10−19−6.63×10−206.626×10−34=9.99×1013=1014 s−1

The maximum kinetic energy of the photoelectrons ejected from a metal when it is irradiated with a radiation of frequency $2 \times 10^{14} s^{- 1}$ is $6.63 \times 10^{- 20} J$ . The thereshold frequency of the metal is: