1 attometer is ………. Nanometre
(A) \[{10^{ - 9}}\]
(B) \[{10^{ - 8}}\]
(C) \[{10^{ - 7}}\]
(D) \[{10^9}\]

Hint: 1 attometre is equal to \[1 \times {10^{ - 18}}\] metres. 1 nanometer is equal to \[1 \times {10^{ - 9}}\] meters. Using the two relations, we can multiply or divide them using the unit as a guide.

Complete Step-by-Step solution:
By the phrasal of the question, we are to find the equivalent of 1 attometre (a unit of length) in nanometre. To do this, we can use the knowledge of their conversion to metres. What is meant by this is to use the knowledge of the equivalent of attometre to metre, and the equivalent of 1 nanometre to metres.
1 attometre is equal to \[1 \times {10^{ - 18}}\] metres i.e. \[1am = {10^{ - 18}}m\] and this also implies that \[{10^{18}}\] attometre makes 1 metre that is \[{10^{18}}am = 1m\]. This can be written as \[{10^{18}}\dfrac{{am}}{m}\]
1 nanometre is equal to \[1 \times {10^{ - 9}}\] metres i.e. \[1nm = {10^{ - 9}}m\] and this can be written as \[{10^{ - 9}}\dfrac{m}{{nm}}\]
Hence to find the equivalent of attometre in nanometre, we shall do as follows
\[{10^{ - 9}}\dfrac{m}{{nm}} \times {10^{18}}\dfrac{{am}}{m}\]
Hence, by computation and cancellation of \[m\], we have
\[{10^9}\dfrac{{am}}{{nm}}\]
Then this means that \[{10^9}\] is equal to 1 nm.
Thus, by inverting \[{10^9}\dfrac{{am}}{{nm}}\], we have
\[\dfrac{1}{{{{10}^9}}}\dfrac{{nm}}{{am}}\] which is equivalent to \[{10^{ - 9}}\dfrac{{am}}{{nm}}\]. This implies that the \[{10^{ - 9}}\] nanometre makes 1 attometer

Hence, the correct option is A.

Note: Alternatively, we could reason as follows, if
\[{10^{ - 18}}\] am is 1 m, then \[{10^{ - 9}}\] m would be equal to
\[\dfrac{{{{10}^{ - 9}}}}{{{{10}^{ - 18}}}}\] am, and this is equal to \[{10^9}\]am.
Now, but \[{10^{ - 9}}\] m is 1 nm. Then \[{10^9}\] am is actually equal to 1 nm.
Then 1 am is equal \[\dfrac{1}{{{{10}^9}}}\]am. And this is equal to \[{10^{ - 9}}\]nm.
Then 1 attometre is, indeed, equal to \[{10^{ - 9}}\]nm, which is identical to as calculated in step by step solution.