Question

The threshold frequency for a certain metal is 3.3 X 10¹⁴ Hz. If light of frequency 8.2 X 10¹⁴ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.​

Answer 1

Answer:

Explanation:

The threshold frequency of metal is 3.3×10

14

 Hz.

Frequency of the light incident on metal is 8.2×10

14

 Hz.

To find: The cut-off voltage for photoelectric emission.

The cut-off voltage is given by:

eV

o

​

=h(ν−ν

o

​

)

The cut-off voltage is V

o

​

.

Here, ν

o

​

=3.3×10

14

Hz and ν=8.2×10

14

Hz

So,

V

0

​

=

1.6×10

−19

6.6×10

−34

(8.2×10

14

−3.3×10

14

)

​

⇒2.012 V

Answer 2

AnSwer :

Threshold frequency of the metal v$_0$ = 3.3 × 10¹⁴Hz.

frequency of light incident on the metal, v = 8.2 × 10¹⁴Hz

Charge on an electron, e = 1.6 × 10¯19C.

Planck's constant, h = 6.626 × 10¯³⁴Js.

cut-off voltage for the photoelectric emissions from the metal = v$_0$

The equation for the cut-off energy is given as,

${ \leadsto{ \bf{eV_0 = (v - v_0)}}}$

${ \leadsto{ \bf{V_0 = \dfrac{h(v -v_0) }{e} }}}$

${ \leadsto{ \bf{V_0 = \dfrac{6.626 \times {10}^{ - 34} \times (8.2 \times {10}^{14} - 3.3 \times {10}^{14} ) }{1.6 \times {10}^{ - 19} } }}}$

${ \leadsto{ \bf{V_0 = 2.0292v }}}$

thus, the cut-off voltage for the photoelectric emissions is 2.0292v.